Ohanian Capitulo 20

CHAPTER 20

HEAT

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored.
†20-1. Since no heat is lost to the surroundings, the temperature of the water will increase as heat Q is added. The amount of temperature increase ∆T is determined by the mass of the water m and itsspecific heat c. The heater converts electricity at a rate P = 620 W to increase the temperature of the water. This power P is equal to a rate of 620 J/s. To determine the time required to increase the temperature of the water, we recognize that Q mc∆T P= = t t
1.0 kg mc∆T (1.0 liter ) 1.0 liter ( 4187 J/(kg i °C) )( 80 °C ) = = 540 s P 620 J/s Assume each child converts 2000 kcal of food to heatthroughout the day. The power associated with 1000 children in the school for seven hours is kcal 2000 child ( 4187 J ) Q 1 kcal P= = 1000 children × = 3.3 × 105 W or 330 kW t 7 h ( 3600 s ) 1h

t=

(

)

20-2.

†20-3.

Assume that all of the gravitational potential energy (mgh) of the water is converted into heat Q. Then, the temperature change of the water can be determined by rearrangingthe equation Q = mc∆T. Q mgh gh (9.81 m/s 2 )(120 m) ∆T = = = = = 0.28°C mc mc c 4187 J/(kg i °C)

20-4.

Note that the mass of the water drops out of the equation. Thus, this would be the temperature change regardless of the actual amount of water that fell. Q m P= = c∆T t t P 1200 × 106 J/s ∆T = = = 4.95°C 1 day ( m/t )c (5.0 × 106 m3 /day) 1000 3kg 86 400 s ( 4187 J/(kg i °C) )

(

1m)(

)

20-5.

20-6. †20-7.

Therefore, the final temperature is T = T0 + ∆T = 20°C + 4.95°C = 25°C To burn 100 kcal at a rate of 750 kcal/h requires 100 kcal t= = 0.133 h 750 kcal/h If one is running at a rate of 12 km/h, then the distance run would be x = vt = (12 km/h)(0.133 h) = 1.6 km 2000 kcal ⎛ 4187 J ⎞ ⎛ 1 day ⎞ P= ⎜ ⎟⎜ ⎟ = 97 W 1 day ⎝ 1 kcal ⎠ ⎝ 86 400 s ⎠ The power P you exertrotates the paddles, which, in turn, do mechanical work on the water. The process converts chemical energy in your body into mechanical work and then into a temperature increase of the water.

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P=

20
Q mc∆T = t t

1.0 kg mc∆T 4.0 liters 1.0 liter ( 4186 J/(kg i °C) ) (5.0°C) t= = = 750 s P 0.15 hp 745.7 W ( 1 J/s ) 1 hp 1W

(

)

(

)

20-8.

The temperature ofthe cream must increase after it is added to the coffee. As a result, the mug and coffee must lose heat. Since heat is neither being added nor removed from the system, mcream ccream (T − Tcream ) + mmug cmug (T − T0 ) + mcoffee ccoffee (T − T0 ) = 0 T (mcream ccream + mmug cmug + mcoffee ccoffee ) − mcream ccreamTcream − (mmug cmug + mcoffee ccoffee )T0 = 0
T = = mcream ccreamTcream + ( mmug cmug+ mcoffee ccoffee )T0 mcream ccream + mmug cmug + mcoffee ccoffee (0.015 kg) ( 2900 J/(kg i °C) ) (5 °C) + ⎡ (0.125 kg) ( 840J/(kg i °C) ) + (0.180 kg) ( 4187 J/(kg i °C) ) ⎦ (70°C) ⎤ ⎣ (0.015 kg) ( 2900 J/(kg i °C) ) + (0.125 kg) ( 840J/(kg i °C) ) +(0.180 kg) ( 4187 J/(kg i °C) )

†20-9.

= 67°C Assume that all of the energy contained in the snack is converted by the body into gravitationalpotential energy at a height H above the starting location. Q = mgH = mgnh

20-10.

350 kcal ( 4187 J ) Q 1 kcal = = 8500 steps mgh (70 kg)(9.81 m/s 2 )(0.25 m) In reality, the body is not completely efficient; and it also uses energy for other purposes, so one would have completely used up the energy in the snack in reaching a much lower altitude than indicated. The temperature of the leadplate must decrease after it is dropped into the water. As a result, the lead must lose heat as the water gains it. Since heat is neither being added nor removed from the system, mw cw (T − Tw ) + mlead clead (T − TL ) = 0 n= T (mw cw + mlead clead ) = mw cwTw + mlead cleadTL m c T + mlead cleadTL T = w w w mw cw + mlead clead
= (200 liter)

) ( 4187 J/(kg i °C) ) (25°C) + (20 kg) ( (130 J/(kg...