Content check — University Physics 2, Chapter 1: Temperature and Heat

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 1: Temperature and Heat

74 problems · 0 figures

Check Your Understanding 1.1 §1.3 source ↗

Does a given reading on a gasoline gauge indicate more gasoline in cold weather or in hot weather, or does the temperature not matter?

Answer

The actual amount (mass) of gasoline left in the tank when the gauge hits “empty” is less in the summer than in the winter. The gasoline has the same volume as it does in the winter when the “add fuel” light goes on, but because the gasoline has expanded, there is less mass.

Check Your Understanding 1.2 §1.3 source ↗

Two objects A and B have the same dimensions and are constrained identically. A is made of a material with a higher thermal expansion coefficient than B. If the objects are heated identically, will A feel a greater stress than B?

Answer

Not necessarily, as the thermal stress is also proportional to Young’s modulus.

Check Your Understanding 1.3 §1.4 source ↗

If 25 kJ is necessary to raise the temperature of a rock from

25 °C to 30 ° C,

how much heat is necessary to heat the rock from

45 °C to 50 ° C

Answer

To a good approximation, the heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case. (As we will see in the next section, the answer would have been different if the object had been made of some substance that changes phase anywhere between

30 ° C

and

50 ° C

Check Your Understanding 1.4 §1.5 source ↗

Explain why a cup of water (or soda) with ice cubes stays at

0 ° C,

even on a hot summer day.

Answer

The ice and liquid water are in thermal equilibrium, so that the temperature stays at the freezing temperature as long as ice remains in the liquid. (Once all of the ice melts, the water temperature will start to rise.)

Check Your Understanding 1.5 §1.5 source ↗

Why does snow often remain even when daytime temperatures are higher than the freezing temperature?

Answer

Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be transferred from the air, even if the air is above

0 ° C

Check Your Understanding 1.6 §1.6 source ↗

Name an example from daily life (different from the text) for each mechanism of heat transfer.

Answer

Conduction: Heat transfers into your hands as you hold a hot cup of coffee. Convection: Heat transfers as the barista “steams” cold milk to make hot cocoa. Radiation: Heat transfers from the Sun to a jar of water with tea leaves in it to make “Sun tea.” A great many other answers are possible.

Check Your Understanding 1.7 §1.6 source ↗

How does the rate of heat transfer by conduction change when all spatial dimensions are doubled?

Answer

Because area is the product of two spatial dimensions, it increases by a factor of four when each dimension is doubled

(A_{final} = {(2 d)}^{2} = 4 d^{2} = 4 A_{initial})

. The distance, however, simply doubles. Because the temperature difference and the coefficient of thermal conductivity are independent of the spatial dimensions, the rate of heat transfer by conduction increases by a factor of four divided by two, or two:

P_{final} =

\frac{k A_{final} (T_{h} - T_{c})}{d_{final}} =

\frac{k (4 A_{final} (T_{h} - T_{c}))}{2 d_{initial}} =

2 \frac{k A_{final} (T_{h} - T_{c})}{d_{initial}} =

2 P_{initial}

Check Your Understanding 1.8 §1.6 source ↗

Explain why using a fan in the summer feels refreshing.

Answer

Using a fan increases the flow of air: Warm air near your body is replaced by cooler air from elsewhere. Convection increases the rate of heat transfer so that moving air “feels” cooler than still air.

Check Your Understanding 1.9 §1.6 source ↗

How much greater is the rate of heat radiation when a body is at the temperature

40 ° C

than when it is at the temperature

20 ° C

Answer

The radiated heat is proportional to the fourth power of the absolute temperature. Because

T_{1} = 293 K

and

T_{2} = 313 K

, the rate of heat transfer increases by about 30% of the original rate.

1-1 §1.1 source ↗

What does it mean to say that two systems are in thermal equilibrium?

Answer

They are at the same temperature, and if they are placed in contact, no net heat flows between them.

1-3 §1.2 source ↗

If a thermometer is allowed to come to equilibrium with the air, and a glass of water is not in equilibrium with the air, what will happen to the thermometer reading when it is placed in the water?

Answer

The reading will change.

1-5 §1.3 source ↗

Pouring cold water into hot glass or ceramic cookware can easily break it. What causes the breaking? Explain why Pyrex®, a glass with a small coefficient of linear expansion, is less susceptible.

Answer

The cold water cools part of the inner surface, making it contract, while the rest remains expanded. The strain is too great for the strength of the material. Pyrex contracts less, so it experiences less strain.

1-7 §1.3 source ↗

Does it really help to run hot water over a tight metal lid on a glass jar before trying to open it? Explain your answer.

Answer

In principle, the lid expands more than the jar because metals have higher coefficients of expansion than glass. That should make unscrewing the lid easier. (In practice, getting the lid and jar wet may make gripping them more difficult.)

1-9 §1.3 source ↗

Calculate the length of a 1-meter rod of a material with thermal expansion coefficient

α

when the temperature is raised from 300 K to 600 K. Taking your answer as the new initial length, find the length after the rod is cooled back down to 300 K. Is your answer 1 meter? Should it be? How can you account for the result you got?

Answer

After being heated, the length is (

1 + 300 α

) (

1 m

). After being cooled, the length is

(1 - 300 α) (1 + 300 α) (1 m)

. That answer is not 1 m, but it should be. The explanation is that even if

α

is exactly constant, the relation

Δ L = α L Δ T

is strictly true only in the limit of small

Δ T

. Since

α

values are small, the discrepancy is unimportant in practice.

1-11 §1.4 source ↗

How is heat transfer related to temperature?

Answer

Temperature differences cause heat transfer.

1-13 §1.4 source ↗

When heat transfers into a system, is the energy stored as heat? Explain briefly.

Answer

No, it is stored as thermal energy. A thermodynamic system does not have a well-defined quantity of heat.

1-15 §1.5 source ↗

A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure. How does this greater pressure increase cooking speed?

Answer

It raises the boiling point, so the water, which the food gains heat from, is at a higher temperature.

1-17 §1.5 source ↗

Can carbon dioxide be liquefied at room temperature (

20 ° C

)? If so, how? If not, why not? (See the phase diagram in the preceding problem.)

Answer

Yes, by raising the pressure above 56 atm.

1-19 §1.5 source ↗

Heat transfer can cause temperature and phase changes. What else can cause these changes?

Answer

work

1-21 §1.5 source ↗

What is the temperature of ice right after it is formed by freezing water?

Answer

0 ° C

(at or near atmospheric pressure)

1-23 §1.5 source ↗

What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?

Answer

Condensation releases heat, so it speeds up the melting.

1-25 §1.5 source ↗

In winter, it is often warmer in San Francisco than in Sacramento, 150 km inland. In summer, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures.

Answer

Because of water’s high specific heat, it changes temperature less than land. Also, evaporation reduces temperature rises. The air tends to stay close to equilibrium with the water, so its temperature does not change much where there’s a lot of water around, as in San Francisco but not Sacramento.

1-27 §1.5 source ↗

In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublime, with some crystals lingering for a while and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from Table 1.4.

Table 1.4 Heats of Fusion and Vaporization^[1] in the book ↗

		$L_{f}$			$L_{v}$
Substance	Melting Point $(°C)$	kJ/kg	kcal/kg	Boiling Point $(°C)$	kJ/kg	kcal/kg
Helium^[2]	$-$ $272.2 (0.95 K)$	5.23	1.25	$−268.9 (4.2 K)$	20.9	4.99
Hydrogen	$−259.3 (13.9 K)$	58.6	14.0	$−252.9 (20.2 K)$	452	108
Nitrogen	$−210.0 (63.2 K)$	25.5	6.09	$−195.8 (77.4 K)$	201	48.0
Oxygen	$−218.8 (54.4 K)$	13.8	3.30	$−183.0 (90.2 K)$	213	50.9
Ethanol	–114	104	24.9	78.3	854	204
Ammonia	–75	332	79.3	–33.4	1370	327
Mercury	–38.9	11.8	2.82	357	272	65.0
Water	0.00	334	79.8	100.0	2256^[3]	539^[4]
Sulfur	119	38.1	9.10	444.6	326	77.9
Lead	327	24.5	5.85	1750	871	208
Antimony	631	165	39.4	1440	561	134
Aluminum	660	380	90	2450	11400	2720
Silver	961	88.3	21.1	2193	2336	558
Gold	1063	64.5	15.4	2660	1578	377
Copper	1083	134	32.0	2595	5069	1211
Uranium	1133	84	20	3900	1900	454
Tungsten	3410	184	44	5900	4810	1150

^[1]Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (

1 atm

). ^[2]Helium has no solid phase at atmospheric pressure. The melting point given is at a pressure of 2.5 MPa.^[3] ^[4] At

37.0 ° C

(body temperature), the heat of vaporization

L_{v}

for water is 2430 kJ/kg or 580 kcal/kg.

Answer

The liquid is oxygen, whose boiling point is above that of nitrogen but whose melting point is below the boiling point of liquid nitrogen. The crystals that sublime are carbon dioxide, which has no liquid phase at atmospheric pressure. The crystals that melt are water, whose melting point is above carbon dioxide’s sublimation point. The water came from the instructor’s breath.

1-29 §1.6 source ↗

When our bodies get too warm, they respond by sweating and increasing blood circulation to the surface to transfer thermal energy away from the core. What effect will those processes have on a person in a

40.0 - ° C

hot tub?

Answer

Increasing circulation to the surface will warm the person, as the temperature of the water is warmer than human body temperature. Sweating will cause no evaporative cooling under water or in the humid air immediately above the tub.

1-31 §1.6 source ↗

Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while the surface only a few centimeters away is safe to touch. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?

Answer

It spread the heat over the area above the heating elements, evening the temperature there, but does not spread the heat much beyond the heating elements.

1-33 §1.6 source ↗

One way to make a fireplace more energy-efficient is to have room air circulate around the outside of the fire box and back into the room. Detail the methods of heat transfer involved.

Answer

Heat is conducted from the fire through the fire box to the circulating air and then convected by the air into the room (forced convection).

1-35 §1.6 source ↗

When watching a circus during the day in a large, dark-colored tent, you sense significant heat transfer from the tent. Explain why this occurs.

Answer

The tent is heated by the Sun and transfers heat to you by all three processes, especially radiation.

1-37 §1.6 source ↗

Why are thermometers that are used in weather stations shielded from the sunshine? What does a thermometer measure if it is shielded from the sunshine? What does it measure if it is not?

Answer

If shielded, it measures the air temperature. If not, it measures the combined effect of air temperature and net radiative heat gain from the Sun.

1-39 §1.6 source ↗

Your house will be empty for a while in cold weather, and you want to save energy and money. Should you turn the thermostat down to the lowest level that will protect the house from damage such as freezing pipes, or leave it at the normal temperature? (If you don’t like coming back to a cold house, imagine that a timer controls the heating system so the house will be warm when you get back.) Explain your answer.

Answer

Turn the thermostat down. To have the house at the normal temperature, the heating system must replace all the heat that was lost. For all three mechanisms of heat transfer, the greater the temperature difference between inside and outside, the more heat is lost and must be replaced. So the house should be at the lowest temperature that does not allow freezing damage.

1-41 §1.6 source ↗

Broiling is a method of cooking by radiation, which produces somewhat different results from cooking by conduction or convection. A gas flame or electric heating element produces a very high temperature close to the food and above it. Why is radiation the dominant heat-transfer method in this situation?

Answer

Air is a good insulator, so there is little conduction, and the heated air rises, so there is little convection downward.

1-43 §1.2 source ↗

While traveling outside the United States, you feel sick. A companion gets you a thermometer, which says your temperature is 39. What scale is that on? What is your Fahrenheit temperature? Should you seek medical help?

Answer

That must be Celsius. Your Fahrenheit temperature is

102 ° F .

Yes, it is time to get treatment.

1-45 §1.2 source ↗

(a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when it decreases by

40.0 ° F

? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees

Answer

Δ T_{C} = 22.2 ° C

; b. We know that

Δ T_{F} = T_{F2} - T_{F1}

. We also know that

T_{F2} =

\frac{9}{5} T_{C2} +

32

and

T_{F1} =

\frac{9}{5} T_{C1} +

32 .

So, substituting, we have

Δ T_{F} =

(\frac{9}{5} T_{C2} + 32) -

(\frac{9}{5} T_{C1} + 32)

. Partially solving and rearranging the equation, we have

Δ T_{F} =

\frac{9}{5} (T_{C2} - T_{C1})

. Therefore,

Δ T_{F} = \frac{9}{5} Δ T_{C}

1-47 §1.2 source ↗

(a) At what temperature do the Fahrenheit and Celsius scales have the same numerical value? (b) At what temperature do the Fahrenheit and Kelvin scales have the same numerical value?

Answer

−40 °

; b. 575 K

1-49 §1.3 source ↗

The height of the Washington Monument is measured to be 170.00 m on a day when the temperature is

35.0 ° C.

What will its height be on a day when the temperature falls to

−10.0 ° C

? Although the monument is made of limestone, assume that its coefficient of thermal expansion is the same as that of marble. Give your answer to five significant figures.

Answer

Using Table 1.2 to find the coefficient of thermal expansion of marble:

L =

L_{0} +

Δ L =

L_{0} (1 + α Δ T) =

170 m [1 + (2.5 \times 10^{−6} / ° C) (−45.0 ° C)] =

169.98 m

.
(Answer rounded to five significant figures to show the slight difference in height.)

Table 1.2 Thermal Expansion Coefficients in the book ↗

Material	Coefficient of Linear Expansion $α (1 / ° C)$	Coefficient of Volume Expansion $β (1 / ° C)$
Solids
Aluminum	$25 \times 10^{−6}$	$75 \times 10^{−6}$
Brass	$19 \times 10^{−6}$	$56 \times 10^{−6}$
Copper	$17 \times 10^{−6}$	$51 \times 10^{−6}$
Gold	$14 \times 10^{−6}$	$42 \times 10^{−6}$
Iron or steel	$12 \times 10^{−6}$	$35 \times 10^{−6}$
Invar (nickel-iron alloy)	$0.9 \times 10^{−6}$	$2.7 \times 10^{−6}$
Lead	$29 \times 10^{−6}$	$87 \times 10^{−6}$
Silver	$18 \times 10^{−6}$	$54 \times 10^{−6}$
Glass (ordinary)	$9 \times 10^{−6}$	$27 \times 10^{−6}$
Glass (Pyrex®)	$3 \times 10^{−6}$	$9 \times 10^{−6}$
Quartz	$0.4 \times 10^{−6}$	$1 \times 10^{−6}$
Concrete, brick (approximate)	$12 \times 10^{−6}$	$36 \times 10^{−6}$
Marble (average)	$2.5 \times 10^{−6}$	$7.5 \times 10^{−6}$
Liquids
Ether		$1650 \times 10^{−6}$
Ethyl alcohol		$1100 \times 10^{−6}$
Gasoline		$950 \times 10^{−6}$
Glycerin		$500 \times 10^{−6}$
Mercury		$180 \times 10^{−6}$
Water		$210 \times 10^{−6}$
Gases
Air and most other gases at atmospheric pressure		$3400 \times 10^{−6}$

1-51 §1.3 source ↗

What is the change in length of a 3.00-cm-long column of mercury if its temperature changes from

37.0 ° C

40.0 ° C

, assuming the mercury is constrained to a cylinder but unconstrained in length? Your answer will show why thermometers contain bulbs at the bottom instead of simple columns of liquid.

Answer

We use

β

instead of

α

since this is a volume expansion with constant surface area. Therefore:

Δ L =

α L Δ T =

(6.0 \times 10^{−5} / ° C) (0.0300 m) (3.00 ° C) =

5.4 \times 10^{−6} m

1-53 §1.3 source ↗

You are looking to buy a small piece of land in Hong Kong. The price is “only” $60,000 per square meter. The land title says the dimensions are

20 m \times 30 m

. By how much would the total price change if you measured the parcel with a steel tape measure on a day when the temperature was

20 ° C

above the temperature that the tape measure was designed for? The dimensions of the land do not change.

Answer

On the warmer day, our tape measure will expand linearly. Therefore, each measured dimension will be smaller than the actual dimension of the land. Calling these measured dimensions

l'

and

w'

, we will find a new area, A. Let’s calculate these measured dimensions:

l' =

l_{0} -

Δ l =

(20 m) -

(20 ° C) (20 m) (\frac{1.2 \times 10^{−5}}{° C}) =

19.9952 m

;

A' =

l \times w' =

(29.9928 m) (19.9952 m) =

599.71 m^{2}

;
Cost change

=

(A - A') (\frac{$ 60,000}{m^{2}}) =

((600 - 599.71) m^{2}) (\frac{$ 60,000}{m^{2}}) =

$ 17,000

.
Because the area gets smaller, the price of the land decreases by about $17,000.

1-55 §1.3 source ↗

(a) Suppose a meter stick made of steel and one made of aluminum are the same length at

0^{} ° C

. What is their difference in length at

22.0 ° C

? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes.

Answer

a. Use Table 1.2 to find the coefficients of thermal expansion of steel and aluminum. Then

Δ L_{Al} -

Δ L_{steel} =

(α_{Al} - α_{steel}) L_{0} Δ T =

(\frac{2.5 \times 10^{−5}}{° C} - \frac{1.2 \times 10^{−5}}{° C}) (1.00 m) (22 ° C) =

2.9 \times 10^{−4} m

.
b. By the same method with

L_{0} = 30.0 m

, we have

Δ L =

8.6 \times 10^{−3} m

Table 1.2 Thermal Expansion Coefficients in the book ↗

Material	Coefficient of Linear Expansion $α (1 / ° C)$	Coefficient of Volume Expansion $β (1 / ° C)$
Solids
Aluminum	$25 \times 10^{−6}$	$75 \times 10^{−6}$
Brass	$19 \times 10^{−6}$	$56 \times 10^{−6}$
Copper	$17 \times 10^{−6}$	$51 \times 10^{−6}$
Gold	$14 \times 10^{−6}$	$42 \times 10^{−6}$
Iron or steel	$12 \times 10^{−6}$	$35 \times 10^{−6}$
Invar (nickel-iron alloy)	$0.9 \times 10^{−6}$	$2.7 \times 10^{−6}$
Lead	$29 \times 10^{−6}$	$87 \times 10^{−6}$
Silver	$18 \times 10^{−6}$	$54 \times 10^{−6}$
Glass (ordinary)	$9 \times 10^{−6}$	$27 \times 10^{−6}$
Glass (Pyrex®)	$3 \times 10^{−6}$	$9 \times 10^{−6}$
Quartz	$0.4 \times 10^{−6}$	$1 \times 10^{−6}$
Concrete, brick (approximate)	$12 \times 10^{−6}$	$36 \times 10^{−6}$
Marble (average)	$2.5 \times 10^{−6}$	$7.5 \times 10^{−6}$
Liquids
Ether		$1650 \times 10^{−6}$
Ethyl alcohol		$1100 \times 10^{−6}$
Gasoline		$950 \times 10^{−6}$
Glycerin		$500 \times 10^{−6}$
Mercury		$180 \times 10^{−6}$
Water		$210 \times 10^{−6}$
Gases
Air and most other gases at atmospheric pressure		$3400 \times 10^{−6}$

1-57 §1.3 source ↗

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at

10.0 ° C

. What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of

95.0 ° C,

given that the fluid’s volume coefficient of expansion is

β =

400 \times 10^{−6} / ° C

? (Your answer will be a conservative estimate, as most car radiators have operating temperatures greater than

95.0 ° C

Answer

Δ V = 0.475 L

1-59 §1.3 source ↗

The density of water at

0 ° C

is very nearly

1000 {kg/m}^{3}

(it is actually

999.84 {kg/m}^{3}

), whereas the density of ice at

0 ° C

917 {kg/m}^{3} .

Calculate the pressure necessary to keep ice from expanding when it freezes, neglecting the effect such a large pressure would have on the freezing temperature. (This problem gives you only an indication of how large the forces associated with freezing water might be.)

Answer

If we start with the freezing of water, then it would expand to

(1 m^{3}) (\frac{1000 {kg/m}^{3}}{917 {kg/m}^{3}}) =

1.09 m^{3} =

1.98 \times 10^{8} {N/m}^{2}

of ice.

1-61 §1.4 source ↗

On a hot day, the temperature of an 80,000-L swimming pool increases by

1.50 ° C

. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.

Answer

m = 5.02 \times 10^{8} J

1-63 §1.4 source ↗

The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at

20.0 ° C

: (a) water; (b) concrete; (c) steel; and (d) mercury.

Answer

Q =

m c Δ T \Rightarrow Δ T =

\frac{Q}{m c}

; a.

21.0 ° C

; b.

25.0 ° C

; c.

29.3 ° C

; d.

50.0 ° C

1-65 §1.4 source ↗

0.250 -kg

block of a pure material is heated from

20.0 ° C

65.0 ° C

by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.

Answer

Q =

m c Δ T \Rightarrow c =

\frac{Q}{m Δ T} =

\frac{1.04 kcal}{(0.250 kg) (45.0 ° C)} =

0.0924 kcal/kg \cdot ° C

. It is copper.

1-67 §1.4 source ↗

(a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a

54.9 - ° C

temperature increase? Assume the process takes place in an ideal calorimeter, in other words a perfectly insulated container. (b) Compare your answer to the following labeling information found on a package of dry roasted peanuts: a serving of 33 g contains 200 calories. Comment on whether the values are consistent.

Answer

Q =

m_{w} c_{w} Δ T +

m_{A1} c_{A1} Δ T =

(m_{w} c_{w} + m_{A1} c_{A1}) Δ T

;

Q = [\begin{matrix} (0.500 kg) (1.00 kcal/kg \cdot ° C) + \\ (0.100 kg) (0.215 kcal/kg \cdot ° C) \end{matrix}] (54.9 ° C) = 28.63 kcal

;

\frac{Q}{m_{p}} =

\frac{28.63 kcal}{5.00 g} =

5.73 kcal/g

; b.

\frac{Q}{m_{p}} =

\frac{200 kcal}{33 g} =

6 kcal/g,

which is consistent with our results to part (a), to one significant figure.

1-69 §1.4 source ↗

In a study of healthy young men¹, doing 20 push-ups in 1 minute burned an amount of energy per kg that for a 70.0-kg man corresponds to 8.06 calories (kcal). How much would a 70.0-kg man’s temperature rise if he did not lose any heat during that time?

Answer

0.139 ° C

1-71 §1.4 source ↗

Repeat the preceding problem, assuming the water is in a glass beaker with a mass of 0.200 kg, which in turn is in a calorimeter. The beaker is initially at the same temperature as the water. Before doing the problem, should the answer be higher or lower than the preceding answer? Comparing the mass and specific heat of the beaker to those of the water, do you think the beaker will make much difference?

Answer

It should be lower. The beaker will not make much difference:

16.3 ° C

1-73 §1.5 source ↗

A bag containing

0 ° C

ice is much more effective in absorbing energy than one containing the same amount of

0 ° C

water. (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from

0 ° C

30.0 ° C

? (b) How much heat transfer is required to first melt 0.800 kg of

0 ° C

ice and then raise its temperature to

30.0 ° C

? (c) Explain how your answer supports the contention that the ice is more effective.

Answer

1.00 \times 10^{5} J

; b.

3.68 \times 10^{5} J

; c. The ice is much more effective in absorbing heat because it first must be melted, which requires a lot of energy, and then it gains the same amount of heat as the bag that started with water. The first

2.67 \times 10^{5} J

of heat is used to melt the ice, then it absorbs the

1.00 \times 10^{5} J

of heat as water.

1-75 §1.5 source ↗

Condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of vapor condense on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs. Use

L_{v}

for water at

37 ° C

as a better approximation than

L_{v}

for water at

100 ° C

Answer

58.1 g

1-77 §1.5 source ↗

On a certain dry sunny day, a swimming pool’s temperature would rise by

1.50 ° C

if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant?

Answer

Let M be the mass of pool water and m be the mass of pool water that evaporates.

M c Δ T =

m L_{V (37 ° C)} \Rightarrow \frac{m}{M} =

\frac{c Δ T}{L_{V (37 ° C)}} =

\frac{(1.00 kcal/kg \cdot ° C) (1.50 ° C)}{580 kcal/kg} =

2.59 \times 10^{−3}

;
(Note that

L_{V}

for water at

37 ° C

is used here as a better approximation than

L_{V}

for

100 ° C

water.)

1-79 §1.5 source ↗

In 1986, an enormous iceberg broke away from the Ross Ice Shelf in Antarctica. It was an approximately rectangular prism 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is

917 {kg/m}^{3}

? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of

100 {W/m}^{2}

, 12.00 h per day?

Answer

1.47 \times 10^{15} kg

; b.

4.90 \times 10^{20} J

; c. 48.5 y

1-81 §1.5 source ↗

(a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases

2.80 \times 10^{7} J

of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water’s temperature rises from

20.0 ° C

100 ° C

, it boils, and the resulting steam’s temperature rises to

300 ° C

at constant pressure. (b) Discuss additional complications caused by the fact that crude oil is less dense than water.

Answer

a. 9.35 L; b. Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less able to absorb the heat generated by the oil.

1-83 §1.5 source ↗

To help prevent frost damage, 4.00 kg of water at

0 ° C

is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be

3.35 kJ/kg \cdot ° C

, and assume that no phase change occurs in the tree.

Answer

a. 319 kcal; b.

2.00 ° C

1-85 §1.5 source ↗

A 0.0500-kg ice cube at

−30.0 ° C

is placed in 0.400 kg of

35.0 - ° C

water in a very well-insulated container. What is the final temperature?

Answer

First bring the ice up to

0 ° C

and melt it with heat

Q_{1} :

4.74 kcal. This lowers the temperature of water by

Δ T_{2} :

23.15 ° C

. Now, the heat lost by the hot water equals that gained by the cold water (

T_{f}

is the final temperature):

20.6 ° C

1-87 §1.5 source ↗

Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of

500 - ° C

granite must be placed in 4.00 kg of

15.0 - ° C

water to bring its temperature to

100 ° C

, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings.

Answer

Let the subscripts r, e, v, and w represent rock, equilibrium, vapor, and water, respectively.

m_{r} c_{r} (T_{1} - T_{e}) =

m_{V} L_{V} +

m_{W} c_{W} (T_{e} - T_{2})

;

\begin{matrix} m_{r} & = \frac{m_{V} L_{V} + m_{W} c_{W} (T_{e} - T_{2})}{c_{r} (T_{1} - T_{e})} \\ = \frac{(0.0250 kg) (2256 \times 10^{3} J/kg) + (3.975 kg) (4186 \times 10^{3} J/kg \cdot ° C) (100 ° C - 15 ° C)}{(840 J/kg \cdot ° C) (500 ° C - 100 ° C)} \\ = 4.38 kg \end{matrix}

1-89 §1.6 source ↗

(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The walls’ surface area is

120 m^{2}

and their inside surface is at

18.0 ° C

, while their outside surface is at

5.00 ° C

. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?

Answer

1.01 \times 10^{3} W

; b. One 1-kilowatt room heater is needed.

1-91 §1.6 source ↗

Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is

37.0 ° C

, the skin temperature is

34.0 ° C

, the thickness of the fatty tissues between the core and the skin averages 1.00 cm, and the surface area is

1.40 m^{2}

Answer

84.0 W

1-93 §1.6 source ↗

A man consumes 3000 kcal of food in one day, converting most of it to thermal energy to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?

Answer

2.59 kg

1-95 §1.6 source ↗

(a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a

1.40 {-m}^{2}

surface area? Assume that the animal’s skin temperature is

32.0 ° C

, that the air temperature is

−5.00 ° C

, and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?

Answer

a. 39.7 W; b. 820 kcal

1-97 §1.6 source ↗

Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of

10.0 m^{2}

and a thermal conductivity twice that of glass wool with the rate of heat conduction through a 0.750-cm-thick window that has an area of

2.00 m^{2}

, assuming the same temperature difference across each.

Answer

\frac{Q}{t} =

\frac{k A (T_{2} - T_{1})}{d}

, so that

\frac{{(Q / t)}_{wall}}{{(Q / t)}_{window}} =

\frac{k_{wall} A_{wall} d_{window}}{k_{window} A_{window} d_{wall}} =

\frac{(2 \times 0.042 J/s \cdot m \cdot ° C) (10.0 m^{2}) (0.750 \times 10^{−2} m)}{(0.84 J/s \cdot m \cdot ° C) (2.00 m^{2}) (13.0 \times 10^{−2} m)}

This gives 0.0288 wall: window, or 35:1 window: wall

1-99 §1.6 source ↗

Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through a 14 cm radius and at a rate of 2256 W, what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick.

Answer

\frac{Q}{t} =

\frac{k A (T_{2} - T_{1})}{d} =

\frac{k A Δ T}{d} \Rightarrow

Δ T =

\frac{d (Q / t)}{k A} =

\frac{(6.00 \times 10^{−3} m) (2256 W)}{(0.84 J/s \cdot m \cdot ° C) (0.0616 m^{2})} =

262 K

1-101 §1.6 source ↗

Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as 2 years.) Suppose you wish to install the extra insulation in the preceding problem. If energy cost

$ 1.00

per million joules and the insulation was $4.00 per square meter, then calculate the simple payback time. Take the average

Δ T

for the 120-day heating season to be

15.0 ° C .

Answer

We found in the preceding problem that

P = 126 Δ T W \cdot ° C

as baseline energy use. So the total heat loss during this period is

Q =

(126 J/s \cdot ° C) (15.0 ° C) (120 days) (86.4 \times 10^{3} s/day) =

1960 \times 10^{6} J

. At the cost of $1/MJ, the cost is $1960. From an earlier problem, the savings is 12% or $235/y. We need

150 m^{2}

of insulation in the attic. At

$ 4 / m^{2}

, this is a $500 cost. So the payback period is

$ 600 / ($ 235 / y) =

2.6 years

(excluding labor costs).

1-103 §1.AP source ↗

What is the percent error of thinking the melting point of tungsten is

3695 ° C

instead of the correct value of 3695 K?

Answer

7.39 %

1-105 §1.AP source ↗

How much stress is created in a steel beam if its temperature changes from

–15 ° C

40 ° C

but it cannot expand? For steel, the Young’s modulus

Y =

210 \times 10^{9} {N/m}^{2}

from Stress, Strain, and Elastic Modulus. (Ignore the change in area resulting from the expansion.)

Answer

\frac{F}{A} =

(210 \times 10^{9} Pa) (12 \times 10^{−6} / ° C) (40 ° C - (−15 ° C)) =

1.4 \times 10^{8} {N/m}^{2}

1-107 §1.AP source ↗

A mercury thermometer still in use for meteorology has a bulb with a volume of

0.780 {cm}^{3}

and a tube for the mercury to expand into of inside diameter 0.130 mm. (a) Neglecting the thermal expansion of the glass, what is the spacing between marks

1 ° C

apart? (b) If the thermometer is made of ordinary glass (not a good idea), what is the spacing?

Answer

1.06

cm; b.

0.9

1-109 §1.AP source ↗

You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before you get home so you can eat it at room temperature, the way you like it. Instead, your roommate plays video games for hours. When you return, you notice that the pastry is still cold, but the game console has become hot. Annoyed, and knowing that the pastry will not be good if it is microwaved, you warm up the pastry by unplugging the console and putting it in a clean trash bag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find that the equilibrium temperature is a nice, warm

38.3 ° C

. You know that the game console has a mass of 2.1 kg. Approximate it as having a uniform initial temperature of

45 ° C

. The pastry has a mass of 0.16 kg and a specific heat of

3.0 k J/ (kg \cdot ºC),

and is at a uniform initial temperature of

4.0 ° C

. The plate is at the same temperature and has a mass of 0.24 kg and a specific heat of

0.90 kJ/ (kg \cdot ºC)

. What is the specific heat of the console?

Answer

1.7 kJ/ (kg \cdot ºC)

1-111 §1.AP source ↗

In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of

808 {kg/m}^{3} .

(a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to

3.00 ° C

. (Use

c_{P}

and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of

0 - ° C

ice with that from evaporating the liquid nitrogen.

Answer

1.57 \times 10^{4} kcal

; b.

18.3 kW \cdot h

; c.

1.29 \times 10^{4} kcal

1-113 §1.AP source ↗

A 0.800-kg iron cylinder at a temperature of

1.00 \times 10^{3} ° C

is dropped into an insulated chest of 1.00 kg of ice at its melting point. What is the final temperature, and how much ice has melted?

Answer

6.3 ° C

. All of the ice melted.

1-115 §1.AP source ↗

Repeat the preceding problem with 0.500 kg of ice, assuming that the ice is initially in a copper container of mass 1.50 kg in equilibrium with the ice.

Answer

63.9 ° C

, all the ice melted

1-117 §1.AP source ↗

(a) Calculate the rate of heat conduction through a double-paned window that has a

1.50 {-m}^{2}

area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside surface temperature is

15.0 ° C,

while that on the outside is

−10.0 ° C .

(Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).

Answer

a. 83 W; b.

1.97 \times 10^{3} W

; The single-pane window has a rate of heat conduction equal to 1969/83, or 24 times that of a double-pane window.

1-119 §1.AP source ↗

For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the difference in temperature is

2.00 ° C

, and the skin area is

1.50 m^{2}

. How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.)

Answer

The rate of heat transfer by conduction is 20.0 W. On a daily basis, this is 1,728 kJ/day. Daily food intake is

2400 kcal/d \times 4186 J/kcal =

10,050 kJ/day

. So only 17.2% of energy intake goes as heat transfer by conduction to the environment at this

Δ T

1-121 §1.AP source ↗

An infrared heater for a sauna has a surface area of

0.050 m^{2}

and an emissivity of 0.84. What temperature must it run at if the required power is 360 W? Neglect the temperature of the environment.

Answer

620 K

1-123 §1.CP source ↗

A pendulum is made of a rod of length L and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by dT, the period of the pendulum increases by a fraction

α d T / 2

. (b) A clock controlled by a brass pendulum keeps time correctly at

10 ° C

. If the room temperature is

30 ° C

, does the clock run faster or slower? What is its error in seconds per day?

Answer

Denoting the period by P, we know

P = 2 π \sqrt{L / g} .

When the temperature increases by dT, the length increases by

α L d T

. Then the new length is a.

P =

2 π \sqrt{\frac{L + α L d T}{g}} =

2 π \sqrt{\frac{L}{g} (1 + α d T)} =

2 π \sqrt{\frac{L}{g}} (1 + \frac{1}{2} α d T) =

P (1 + \frac{1}{2} α d T)

by the binomial expansion. b. The clock runs slower, as its new period is 1.00019 s. It loses 16.4 s per day.

1-125 §1.CP source ↗

In a calorimeter of negligible heat capacity, 200 g of steam at

150 ° C

and 100 g of ice at

−40 ° C

are mixed. The pressure is maintained at 1 atm. What is the final temperature, and how much steam, ice, and water are present?

Answer

The amount of heat to melt the ice and raise it to

100 ° C

is not enough to condense the steam, but it is more than enough to lower the steam’s temperature by

50 ° C

, so the final state will consist of steam and liquid water in equilibrium, and the final temperature is

100 ° C

; 28 g of steam condenses, so the final state contains 172 g of steam and 128 g of liquid water.

1-127 §1.CP source ↗

Find the growth of an ice layer as a function of time in a Dewar flask as seen in Exercise 1.120. Call the thickness of the ice layer L. (a) Derive an equation for dL/dt in terms of L , the temperature T above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at

t = 0

, you have

L = 0 .

If you have studied differential equations, you will know a technique for solving equations of this type: manipulate the equation to get dL/dt multiplied by a (very simple) function of L on one side, and integrate both sides with respect to time. Alternatively, you may be able to use your knowledge of the derivatives of various functions to guess the solution, which has a simple dependence on t. (c) Will the water eventually freeze to the bottom of the flask?

Answer

d L / d T = k T / ρ L

; b.

L = \sqrt{2 k T t / ρ L_{f}}

; c. yes

1-129 §1.CP source ↗

Let’s stop ignoring the greenhouse effect and incorporate it into the previous problem in a very rough way. Assume the atmosphere is a single layer, a spherical shell around Earth, with an emissivity $e = 0.77$ (chosen simply to give the right answer) at infrared wavelengths emitted by Earth and by the atmosphere. However, the atmosphere is transparent to the Sun’s radiation (that is, assume the radiation is at visible wavelengths with no infrared), so the Sun’s radiation reaches the surface. The greenhouse effect comes from the difference between the atmosphere’s transmission of visible light and its rather strong absorption of infrared. Note that the atmosphere’s radius is not significantly different from Earth’s, but since the atmosphere is a layer above Earth, it emits radiation both upward and downward, so it has twice Earth’s area. There are three radiative energy transfers in this problem: solar radiation absorbed by Earth’s surface; infrared radiation from the surface, which is absorbed by the atmosphere according to its emissivity; and infrared radiation from the atmosphere, half of which is absorbed by Earth and half of which goes out into space. Apply the method of the previous problem to get an equation for Earth’s surface and one for the atmosphere, and solve them for the two unknown temperatures, surface and atmosphere.

(a) In terms of Earth’s radius, the constant $σ$ , and the unknown temperature $T_{s}$ of the surface, what is the power of the infrared radiation from the surface?

(b) What is the power of Earth’s radiation absorbed by the atmosphere?

(d) Write an equation that says the power of the radiation the atmosphere absorbs from Earth equals the power of the radiation it emits.

(e) Half of the power radiated by the atmosphere hits Earth. Write an equation that says that the power Earth absorbs from the atmosphere and the Sun equals the power that it emits.

(f) Solve your two equations for the unknown temperature of Earth.

Answer

4 (π R^{2}) T_{s}^{4}

; b.

4 e σ π R^{2} T_{s}^{4}

; c.

8 e σ π R^{2} T_{e}^{4}

; d.

T_{s}^{4} = 2 T_{e}^{4}

; e.

(e σ T_{s}^{4}) / 2 +

((1 - A) S) / 4 =

σ T_{s}^{4}

; f.

288 K

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Problems and answers are adapted from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.