# Xxxxxfxxfxx

## Xxxxxfxxfxx

Problem
21. How much heat is required to raise an 800-g
copper pan from 15±C to 90±C if (a) the pan is
empty; (b) the pan contains 1.0 kg of water;
(c) the pan contains 4.0 kg of mercury?
Solution
(a) When just the pan is heated, ¢Q = mCucCu ¢T =
(0:8 kg)(386 J=kg¢K)(90 ¡ 15)K = 23:2 kJ = 5:54 kcal:

Problem
22. Initially, 100 g of water and 100 g of another
substance listed in Table 19-1 are at 20±C: Heat is
then transferred to each substance at the same
rate for 1.0 min. At the end of that time the water
is at 32±C and the other substance at 76±C:
(a) What is the other substance? (b) What is the
heating rate?
Solution
(a) Since the heat energy transferred to both
substances is the same, ¢Q = mWcW¢TW =
mxcx¢Tx; or cx = cW(mW=mx)(¢TW=¢Tx) =
(4184 J=kg¢K)(1)(32 ¡ 20)=(76 ¡ 20) = 897 J=kg¢K:
This is nearest the value for aluminium in Table 19-1.
(b) The rate of heating is ¢Q=¢t = mWcW£
¢TW=¢t = (0:1 kg)(4184 J=kg¢K)(12 K)=(60 s) =
83:7 W = 1:20 kcal=min:

Problem
29. A 1.2-kg iron tea kettle sits on a 2.0-kW stove
burner. If it takes 5.4 min to bring the kettle and
the water in it from 20±C to the boiling point,
how much water is in the kettle?
Solution
The energy supplied by the stove burner heats the
kettle and water in it from 20±C to 100±C. If we
neglect any losses of heat and the heat capacity of the
burner, this energy is just the burner’s power output
times the time, so ¢Q = P ¢t = (mWcW + mKcK)£
¢T (Equation 19-5 for water and kettle). Since all of
these quantities are given except for the mass of the
water, we can solve for mW :
mW = [(P ¢t=¢T) ¡ mKcK]=cW
= f[(2 kW)(5:4 £ 60 s)=80 K] ¡ (1:2 kg)
£(447 J=kg¢K)g=(4184 J=kg¢K)
= 1:81 kg:

Problem
34. A child complains that her cocoa is too hot. The
cocoa is at 90±C: Her father pours 2 oz of milk at
3±C into the 6 oz of cocoa. Assuming milk and
cocoa have the same specific heat as water, what
is the new temperature of the cocoa?
Solution
If the specific heats...