PART
3
THERMODYNAMICS
CHAPTER
19
TEMPERATURE AND
HEAT





Problem
Problem
3. Normal room temperature is 68OF. What
is this in
Celsius?
Solution
Equation 193, solved for the Celsius temperature,
gives Tc
=
i(TF

32)
=
5(68

32)/9
=
20°C.
Sections 194 and 195:
Temperature and
Heat, Heat Capacity and Specific Heat
Problem
14. The average human diet contains 2000 kcal per
day. If all this food energy is released rather than
being stored as fat, what is the approximate
average power output of the human body?
Solution
pa,
=
(2x10~
cal/d)(4.184 J/cal)(l d/86,400 S)
=
96.9 W.
Problem
19.
A
circular lake 1.0 krn in diameter averages 10 m
deep. Solar energy is incident on the lake at an
average rate of 200 W/m2. If the lake absorbs all
this energy and does not exchange heat with its
surroundings, how long will it take to warm from
10°C to 20°C?
27.
A
stove burner supplies heat at the rate of 1.0 kW,
a microwave oven at 625 W. You can heat water in
the microwave in a paper cup of negligible heat
capacity, but the stove requires a pan whose heat
capacity is 1.4 kJ/K. (a) How much water do you
need before it becomes quicker to heat on the
stovetop? (b) What will be the rate at which the
temperzture of this much water rises?
Solution
The temperature rise per second is equal to the heat
supplied per second (i.e., the power supplied if there
are no losses) divided by the total heat capacity of the
water and its container: AT/At
=
(AQ/At)/Cto,
=
P/Ctot (Equation 194 divided by At). The total heat
capacity is Ctot
=
CW
+
Cent, provided the water and
container both have the same instantaneous
temperature. (This assumes that heat is supplied
sufficiently slowly that the water and container share
it and stay in instantaneous thermal equilibrium.) For
the paper cup used in the microwave oven, C,,,
x
0,
whereas for the pan used on the stove burner,
Cent
=
1.4
kJ/K.
Thus, the rate of temperature rise
is

625 W/Cw for the microwave and 1.0 kW/(Cw
+
1.4 kJ/K) for the stove burner. (a) When
=
mwcw is small, the microwave is faster, whereas
when Cw is large, the stove burner is faster. (To see
this, plot both rates as a function of
Cw.)
The rates
of temperature rise are equal for Cw
=
mw
x
(4.184 kJ1kg.K)
=
(1.4 kJ/K)(0.625)/(1

0.625)
=
2.33 kJ/K. Therefore, mw
=
(2.33 kJ/K)+
Solution
(4.184 kJ/kgK)
=
0.558 kg. (b) For mw in part (a),
Since the energy absorbed by ttw lake equals the solar
the rate of temperature rise is ATlAt
=
(0.625 kW)s
power times the time, At
=
AQ/P
=
,CAT+
(2.33 kJ/K)
=
1.0 kWl(2.33