- Submitted By: shahrukh11
- Date Submitted: 02/22/2009 1:25 PM
- Category: American History
- Words: 669
- Page: 3
- Views: 5

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...