heat engine

Let M be the mass of coal burnt in 1 second. Hence, heat input to the first engine = 2.8 x 10^7M J
Efficiency of first engine = 0.6 x [1- (273+440)/(273+670)] = 0.1463
hence, mechanical work output = 0.1463 x 2.8x10^7M J = 4.098 x 10^6M J

Heat released by the first engine = 2.8x10^7M - 4.098x10^6M J
= 2.39 x 10^7M J

Efficiency of second engine = 0.6 x [1- (273+290)/(273+430)] = 0.119
Assume all heat released by the first engine is input into the second engine,
mechanical work ouput from the second engine = 0.119 x 2.39x10^7M J
= 2.86 x 10^6M J
Total mechanical output from the two engines
= (4.098 x 10^6M + 2.86 x 10^6M ) J = 6.958x10^6M J

Hence, 1100 = 6.958M
i.e. M = 1100/6.958 kg/s = 158 kg/s