Question

In refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electrical motor. If the motor is of 1 kW power and heat is transferred from -3°C to 27°C. find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

Answer 1

Answer:

The heat taken out of the refrigerator is 19 kJ

Explanation:

No possible machine is 100% perfect. Carnot machine has the possibility of working between two temperatures with high efficiency.

The efficiency of the engine is given by the formula

E of Eng. $=1-\frac{T 2}{T 1}=1-\frac{270}{300}=0.1$

Therefore the efficiency of the refrigerator is 50% of 0.1 which is  

E of ref  $\frac{50}{100} \times 0.1=0.05$

Since, power = 1 kW hence work done is 1 kJ in 1sec

Therefore, the heat taken out is  

W/ (E of ref) $-1=\frac{1 k J}{0.05}-1$

= 20 - 1 = 19 kJ