The expressions that apply to the treatment of refrigerators (Problem 3A.4) also describe the behaviour of heat pumps, where warmth is obtained from the back of a refrigerator while its front is being used to cool the outside world. Heat pumps are popular home heating devices because they are very efficient. Compare heating of a room at 295K by each of two methods:

(a) Direct conversion of 1.00 kJ of electrical energy in an electrical heater, 

(b) Use of 1.00 kJ of electrical energy to run a reversible heat pump with the outside at 260K. Discuss the origin of the difference in the energy delivered to the interior of the house by the two methods.

Data in (Problem 3A.4)

To calculate the work required to lower the temperature of an object, we need to consider how the coefficient of performance c (see Impact I3.1) changes with the temperature of the object.

(a) Find an expression for the work of cooling an object from T_i to T_f when the refrigerator is in a room at a temperature T_h. Write dw=dq/c(T), relate dq to dT through the heat capacity C_p, and integrate the resulting expression. Assume that the heat capacity is independent of temperature in the range of interest.

(b) Use the result in part (a) to calculate the work needed to freeze 250 g of water in a refrigerator at 293K. How long will it take when the refrigerator operates at 100W?