Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be: [Take specific heat of water = 1 cal g⁻¹ °C⁻¹ and latent heat of steam = 540 cal g⁻¹]

Question

Accepted Answer

31.5 g. The heat lost by steam during condensation and cooling equals the heat gained by water to reach 80°C. Let m be the mass of condensed steam. Heat lost by steam = m × 540 + m × 1 × (100 - 80). Heat gained by water = 20 × 1 × (80 - 10). Equating, m × 540 + m × 20 = 20 × 70. Solving, m = 11.5 g. Total water mass = 20 + 11.5 = 31.5 g.

Steam at 100°C is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be: [Take specific heat of water = 1 cal g⁻¹ °C⁻¹ and latent heat of steam = 540 cal g⁻¹]

Solution

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