Consider an ideal vapour compression refrigeration cycle working on R-134a refrigerant. The COP of the cycle is 10 and the refrigeration capacity is 150 kJ/kg. The heat rejected by the refrigerant in the condenser is __________ kJ/kg (round off to the nearest integer).

The correct answer is 165.

To understand why this is the correct answer, let us break down the thermodynamic principles of the ideal vapour compression refrigeration cycle step-by-step.

1. Understanding the Given Parameters:
We are given the following values for the refrigeration cycle:
Refrigeration capacity (also known as the refrigeration effect), $Q_1=\; 150$ kJ/kg.
Coefficient of Performance of the cycle, $\text{COP}=\; 10$.

2. Calculating the Work Input:
The Coefficient of Performance (COP) of a refrigerator is defined as the ratio of the desired refrigeration effect to the net work input required by the compressor:

$\text{COP}=\frac{Q_1}{W_R}$

Here, $W_R$ represents the work input to the compressor per unit mass of refrigerant (in kJ/kg). By rearranging the formula, we can solve for $W_R$:

$W_R{Q}_1\text{COP}$

Substituting the given values into the equation:

$W_{R}15010=\; 15$ kJ/kg

3. Determining the Heat Rejected in the Condenser:
According to the first law of thermodynamics for a steady-flow cyclic device, the total energy entering the system must equal the total energy leaving the system. For a refrigeration cycle, the heat rejected in the condenser ($Q_2$) is equal to the sum of the heat absorbed in the evaporator ($Q_1$) and the work input to the compressor ($W_R$):

$Q_{2}Q1+W_R$

Substituting the calculated values:

$Q_{2}kJ/kg$

Therefore, the heat rejected by the refrigerant in the condenser is 165 kJ/kg.