A steam engine intakes $$50$$ g of steam at $$100°$$C per minute and cools it down to $$20°$$C. If latent heat of vaporization of steam is $$540$$ cal g$$^{-1}$$, then the heat rejected by the steam engine per minute is ______ $$\times 10^3$$ cal
(Given : specific heat capacity of water : $$1$$ cal g$$^{-1}$$ °C$$^{-1}$$)

The mass of steam is m = 50 g with an initial temperature of 100 °C and a final temperature of 20 °C. The latent heat of vaporization is L = 540 cal/g and the specific heat of water is c = 1 cal/(g·°C).

Condensation of steam to water at 100 °C releases $$Q_1 = mL = 50 \times 540 = 27000 \text{ cal},$$ and cooling this water from 100 °C to 20 °C releases $$Q_2 = mc\Delta T = 50 \times 1 \times (100 - 20) = 4000 \text{ cal}.$$ Therefore, the total heat rejected per minute is $$Q = Q_1 + Q_2 = 27000 + 4000 = 31000 \text{ cal} = 31 \times 10^3 \text{ cal}.$$

The heat rejected by the steam engine per minute is 31 $$\times 10^3$$ cal.