A calorimeter of water equivalent 15 g contains 165 g of water at 25°C. Steam at 100°C is passed through the water for some time. The temperature is…
A calorimeter of water equivalent 15 g contains 165 g of water at 25°C. Steam at 100°C is passed through the water for some time. The temperature is increased to 30°C and the mass of the calorimeter and its contents is increased by 1.5 g. Calculate the specific latent heat of vaporization of water. [Specific heat capacity of water = 4.2 Jg⁻¹K⁻¹]
Water equivalent of the calorimeter 15 g plus 165 g of water gives 180 g of water-equivalent that warms by 5°C (25°C to 30°C).
Heat gained = 180 × 4.2 × 5 = 3780 J.
The mass rose by 1.5 g, so 1.5 g of steam condensed. Heat lost by steam = 1.5L + 1.5 × 4.2 × (100 − 30) = 1.5L + 441.
Set heat lost = heat gained: 1.5L = 3780 − 441 = 3339, so L = 2226 Jg⁻¹.
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