Dry steam at 100°C is bubbled into 20 g of water initially at 10°C. If the final temperature of the water becomes 80°C, what will be the total mass of water present then? [Specific heat capacity of water = 1 cal g⁻¹ °C⁻¹, latent heat of steam = 540 cal g⁻¹]

Question

Accepted Answer

22.5 g. Let m = steam condensed. Heat gained by 20 g water = 20*1*(80-10) = 1400 cal. Heat given by steam = m*540 + m*1*(100-80) = 560m. So m = 1400/560 = 2.5 g, and total mass = 20 + 2.5 = 22.5 g.

Dry steam at 100°C is bubbled into 20 g of water initially at 10°C. If the final temperature of the water becomes 80°C, what will be the total mass of water present then? [Specific heat capacity of water = 1 cal g⁻¹ °C⁻¹, latent heat of steam = 540 cal g⁻¹]

Solution

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