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Question
A vessel of mass 80 g (S.H.C. = 0.8 Jg−1°C−1) contains 250 g of water at 35°C. Calculate the amount of ice at 0°C, which must be added to it, so that the final temperature is 5°C. [Sp. latent heat of ice = 340 Jg−1]
Numerical
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Solution
Let the mass of ice be x g.
Heat gained by ice = 340x + x × 4.2 × 5
= 361x J
Heat lost by water = 250 × 4.2 × (35 − 5)
= 31500 J
Heat lost by vessel = 80 × 0.8 × (35 − 5)
= 1920 J
Total heat lost = 31500 + 1920 = 33420 J
Heat gained = Heat lost
361x = 33420
x = \[\frac{33420}{361}\]
x = 92.57 g
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