Answer to Question #145880 in Classical Mechanics for sridhar

Question

An air conditioner (AC) removes heat at a rate of 1.2kJ/sec from a room. A power of 400W is required to run the AC . The coefficient of performance of the AC is 10% of that of the refrigerator operating between outside and room temperature. If outside temperature is 37°C what will be room temperature?
Ans : 27°C

Answer 1

As per the given question,

Rate of removal of heat =1.2KJ/s = 1200J/s

Power required to run AC = 400W

Coefficient of performance of AC=10%

Outside temperature $T_H=37^\circ$

Room temperature $T_c=?$

We know that coefficient of performance$=1-\frac{T_c}{T_H}$

Now, substituting the values,

$0.1=1-\frac{T_c}{37}$

$\Rightarrow \frac{T_c}{37}=0.9$

$\Rightarrow T_c=0.9\times 37=33.3^\circ$