Heat Pump Work: Find Min Work to Heat Indoor Environment to 27C

[issacnewton]

Hi

I am trying this question from Physics gre prepbook

A heat pump is to extract from an outdoor environment at 7^oC and heat the
environment indoors to 27 celcius. for each 15000 J of heat delivered indoors, the smallest
amount of work that must be supplied to the heat pump is approximately

a)500 J
b)1000 J
c)1100 J
d)2000 J
e) 2200 J

this is how I did my work.

$$T_c = (7+273) K$$

$$T_h=(27+273) K$$

the highest possible efficiency is that of carnot engine

$$\eta_{max}=1-\frac{T_c}{T_h}=\frac{1}{15}$$

let the $\eta$ be the efficiency of this pump . then

$$\eta \leqslant \frac{1}{15}$$

$$\because \; \eta=\frac{W_{eng}}{|Q_h|}$$

$$\therefore \; \frac{W_{eng}}{|Q_h|} \leqslant \frac{1}{15}$$

$$Q_h=15000 J\;\Rightarrow \; W_{eng}\leqslant \frac{15000}{15}$$

$$W_{eng} \leqslant 1000$$

which suggests that W has some maximum value but the problem is talking about the
minimum value of W. is something wrong ?

 

[jbsteele]

No, there is nothing wrong. Since the efficiency of the heat pump is less than the maximum possible efficiency, the minimum work that must be supplied to the heat pump will be greater than 1000 J. The answer is d) 2000 J