Problems and Solutions on Thermodynamics and Statistical Mechanics

36 Problem d Sdutiom on Thermodynamics d Statintical Mechanics

Solution:
(a) The rate of heat from the pump is

At equilibrium, T = T, and Q = Q, = a(Te - To). Thus

(b) In this case, the equilibrium condition is

W = Q(T~ - To).

Thus

W

Ti=To+-<Te.

Q

Therefore it is less desirable than a heat pump.

1040

A room at temperature T2 loses heat to the outside at temperature T1

at a rate A(Tz - Tl). It is warmed by a heat pump operated as a Carnot

cycle between TI and Tz. The power supplied by the heat pump is dW/dT.

(a) What is the maximum rate dQm/dt at which the heat pump can

deliver heat to the room? What is the gain dQm/dW? Evaluate the gain

for tl = 2"C, t2 = 27°C.

(b) Derive an expression for the equilibrium temperature of the room,

(UC, Berkeley)

Tz, in terms of TI, A and dW/dt.

Solution:

(a) From dQm. (TZ - T1)/T2 = dW, we get

With TI = 275K, T2 = 300K, we have dQm/dW = 12.

(b) When equilibrium is reached, one has

T2 dW

A(T2 - Ti) = - -

T2 -TI dt '