**The Second Law of Thermodynamics****Problem 6-55**Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power, determine (a) the COP of the heat pump and (b) the rate of heat absorption from the outside air.**Explanation of Diagram (Figure P6-55):**The diagram illustrates a basic heat pump cycle with the following components:1. **Condenser:** - The refrigerant enters the condenser at 800 kPa and leaves at the same pressure as a saturated liquid. - Heat is rejected from the refrigerant to the surroundings, depicted by \( \dot{Q}_H \).2. **Compressor:** - The refrigerant is compressed, and work is done on the refrigerant by the compressor, denoted by \( \dot{W}_{in} \).3. **Expansion Valve:** - The expansion valve reduces the pressure of the refrigerant, allowing it to evaporate and absorb heat in the next stage.4. **Evaporator:** - The refrigerant absorbs heat from the surroundings, shown by \( \dot{Q}_L \), allowing it to vaporize and complete the cycle.This system is a typical representation of a thermodynamic cycle used in heat pumps, illustrating the energy exchanges involved in heating and cooling processes.

Given, For refrigerant - 134a power consumption by the compressor is, W˙in=1.2kW the pressure at the…

Answered: 6-55 Refrigerant-134a enters the…

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