Answer:
[tex]Q_x=60\ J[/tex]
[tex]T_L=-23\ ^{\circ}C[/tex] is the lowest possible temperature of the cold reservoir.
Explanation:
Given:
Coefficient of performance of refrigerator, [tex]COP=5[/tex]
A)
We know that for a refrigerator:
[tex]\rm COP=\frac{ desired\ effect }{\: energy\ supplied }[/tex]
Given that 10 J of energy is consumed by the compressor:
[tex]5=\frac{desired\ effect}{10}[/tex]
[tex]\rm Desired\ effect=50\ J[/tex]
Now the by the conservation of energy the heat exhausted :
[tex]Q_x=50+10[/tex]
[tex]Q_x=60\ J[/tex]
B)
Also
[tex]COP=\frac{T_L}{T_H-T_L}[/tex]
where:
[tex]T_H\ \&\ T_L[/tex] are the absolute temperatures of high and low temperature reservoirs respectively.
[tex]5=\frac{T_L}{(273+27)-T_L}[/tex]
[tex]1500-5\times T_L=T_L[/tex]
[tex]T_L=250\ K[/tex]
[tex]T_L=-23\ ^{\circ}C[/tex] is the lowest possible temperature of the cold reservoir.