Answer:
a) [tex]T_{H} = 1.967\,^{\circ}F[/tex], b) [tex]COP_{R} = 9.105[/tex], c) [tex]T_{H} = 115.934\,^{\circ}F[/tex], d) [tex]COP_{R} = 6.995[/tex], e) [tex]T_{H} = 25.129\,^{\circ}C[/tex]
Explanation:
a) The coefficient of performance of the reversible refrigeration cycle is:
[tex]COP_{R} = \frac{T_{C}}{T_{H}-T_{C}}[/tex]
[tex]10 = \frac{419.67\,R}{T_{H}-419.67\,R}[/tex]
The temperature of the hot reservoir is:
[tex]10\cdot T_{H} - 4196.7 = 419.67[/tex]
[tex]T_{H} = 461.637\,R[/tex]
[tex]T_{H} = 1.967\,^{\circ}F[/tex]
b) The coefficient of performance is:
[tex]COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}[/tex]
[tex]COP_{R} = 9.105[/tex]
c) The temperature of the hot reservoir can be determined with the help of the following relation:
[tex]\frac{Q_{C}}{Q_{H}-Q_{C}} = \frac{T_{C}}{T_{H}-T_{C}}[/tex]
[tex]\frac{500\,BTU}{600\,BTU-500\,BTU} = \frac{479.67\,R}{T_{H}-479.67\,R}[/tex]
[tex]5 = \frac{479.67\,R}{T_{H}-479.67\,R}[/tex]
[tex]5\cdot T_{H} - 2398.35 = 479.67[/tex]
[tex]T_{H} = 575.604\,R[/tex]
[tex]T_{H} = 115.934\,^{\circ}F[/tex]
d) The coefficient of performance is:
[tex]COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}[/tex]
[tex]COP_{R} = 6.995[/tex]
e) The temperature of the cold reservoir is:
[tex]8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}[/tex]
[tex]8.9\cdot T_{H} - 2386.535 = 268.15[/tex]
[tex]T_{H} = 298.279\,K[/tex]
[tex]T_{H} = 25.129\,^{\circ}C[/tex]
