In the vapour compression cycle shown in the figure, the evaporating and condensing temperature are 260 K and 310 K, respectively. the compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). the specific heat of the liquid refrigerant is 4.8 kJ/kg−K may be treated as constant. The enthalpy of evaporation for the refrigerator at 310 K is 1054 kJ/kg.The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is

In the vapour compression cycle shown in the figure, the evaporating a

1.	In the vapour compression cycle shown in the figure, the evaporating and condensing temperature are 260 K and 310 K, respectively. the compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). the specific heat of the liquid refrigerant is 4.8 kJ/kg−K may be treated as constant. The enthalpy of evaporation for the refrigerator at 310 K is 1054 kJ/kg.The difference between the enthalpies at state points 1 and 0 (in kJ/kg) is
Answer» In the vapour compression cycle shown in the figure, the evaporating and condensing temperature are $260 K$ and $310 K$ , respectively. the compressor takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). the specific heat of the liquid refrigerant is $4.8 k J / k g - K$ may be treated as constant. The enthalpy of evaporation for the refrigerator at $310 K$ is $1054 k J / k g$ . The difference between the enthalpies at state points $1$ and $0$ (in kJ/kg) is