Chapter 4.2, Problem 15P

In an absorption refrigerator the energy driving the process is supplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable.” Let us define the following symbols, all taken to be positive by definition. $\begin{array}{l}{\text{Q}}_{\text{j}}=\text{heat input from flame}\\ {\text{Q}}_{\text{f}}=\text{heat extracted from inside refrigerator}\\ {\text{Q}}_{\text{r}}=\text{waste heat expelled to room}\\ {\text{T}}_{\text{f}}=\text{temperate of flame}\\ {\text{T}}_{\text{c}}=\text{temperature inside refrigerator}\\ {\text{T}}_{\text{r}}=\text{room temperature}\end{array}$

(a) Explain why the “coefficient of performance” (COP) for an absorption refrigerator should be defined as ${\text{Q}}_{\text{c}}/{\text{Q}}_f$.

(b) What relation among ${\text{Q}}_{\text{f}},{\text{Q}}_{\text{c}}$, and ${\text{Q}}_{\text{r}}$ is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1?

(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures ${\text{T}}_{\text{f}},T_c$, and ${\text{T}}_{\text{r}}$ alone.

(d) Explain why a heat pump is better than an electric furnace, which simply converts electrical work directly into heat. (Include some numerical estimates.)