Problems and Solutions on Thermodynamics and Statistical Mechanics

Thermodynum'cs 111

Solution:
(a) The rate that the refrigerator extracts heat from water is

92 = 5 x 320 = 1.6 x lo3 J/s.

The rate that the energy is expelled to the room is

1 - -Q2 Tl = (303/273) x 1.6 x lo3

-T2

= 1.78 x 10 J/s.

(b) The necessary power supplied is

W = Q1- Q2 = 0.18 kW.

(c) The coefficient of performance is

273

&=--- T2 - = 9.1.

Ti - Tz 30

1114

A Carnot cycle is operated with liquid-gas interface. The vapor pres-

sure is pv, temperature T, volume V. The cycle is operated according to

the following p - V diagram.

The cycle goes isothermally from 1 to 2, evaporating n moles of liquid.

This is followed by reversible cooling from 2 to 3, then there is an isother-

mal contraction from 3 to 4, recondensing n moles of liquid, and finally a

reversible heating from 4 to 1, completes the cycle.

I I I

1 pv, T, V Pv-AP pp=p3 --r ,,

b' 1 I I 1 L

"I v2

(a) (b)

Fig. 1.32.

(a) Observe that V2 - V1 = V, - Vt where V, = volume of n moles of

gas, Vt = volume of n moles of liquid. Calculate the efficiency in terms of