Answer:
710.79 Joules
898.56 Joules
264.45 K
172.2 K
Explanation:
W = Work done
Q = Heat added = 1248 J
[tex]T_h[/tex] = Temperature of hot reservoir = 615 K
[tex]T_c[/tex] = Temperature of cold reservoir
Efficiencies of engine A and B
[tex]\eta_A[/tex] = 0.57
[tex]\eta_B[/tex] = 0.72
Efficiency is given by
[tex]\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.57\times 1248\\\Rightarrow W=711.36\ J[/tex]
Work done by engine A is 710.79 Joules
[tex]\eta=\frac{W}{Q}\\\Rightarrow W=\eta Q\\\Rightarrow W=0.72\times 1248\\\Rightarrow W=898.56\ J[/tex]
Work done by engine B is 898.56 Joules
Efficiency
[tex]\eta=1-\frac{T_c}{T_h}\\\Rightarrow 0.57=1-\frac{T_c}{615}\\\Rightarrow T_c=-0.43\times -615\\\Rightarrow T_c=264.45\ K[/tex]
The temperature of the cold reservoir is 264.45 K
[tex]\eta=1-\frac{T_c}{T_h}\\\Rightarrow 0.72=1-\frac{T_c}{615}\\\Rightarrow T_c=-0.28\times -615\\\Rightarrow T_c=172.2\ K[/tex]
The temperature of the cold reservoir is 172.2 K