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 __________ × 10³ cal.
(Given : specific heat capacity of water : 1 cal g⁻¹ °C⁻¹)

Solution

Physics sir · 3 steps

The steam gives up heat twice: first when it condenses, then while the water cools to 20 °C.

1. Condensing. Steam at 100°C first turns into water at 100°C. The heat released is Q₁ = mL = 50 × 540 = 27000 cal.
2. Cooling the water. That water then cools from 100°C to 20°C, a drop of 80°C. The heat released is Q₂ = mcΔ T = 50 × 1 × 80 = 4000 cal.

   Q = mL + mcΔ T = 27000 + 4000 = 31000 cal

3. So. The engine rejects 31 × 10³ cal per minute, so the answer is 31.

The common slip: 27. Counting only the latent heat gives 27. The cooling from 100°C to 20°C adds another 4 × 10³ cal.

The idea to keep. When steam ends up as cooler water, add the latent heat of condensation to the sensible heat of cooling. Use the same mass in both terms.

Answer from the NTA final answer key. Solution by PrepWiser.