A steam engine intakes 50g of steam at 100 ° C per minute and cools...

Heat rejected by the steam engine per minute will be equal to latent heat per minute plus heat rejected by the water to cool down.

$\dfrac{∆ H}{∆ t} = \dfrac{∆ m}{∆ t}L + \dfrac{∆ m}{∆ t}s ∆ T ⇒ \dfrac{∆ H}{∆ t} = 50 \times 540 + 50 \times 1 \times 80 = 31000cal{min}^{- 1} = 31 \times {10}^{3}cal{min}^{- 1}$