Question

A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperatue of water in the beaker is 20ºC. If 440 gm of hot water at 92ºC is poured in it, the final temperature, neglecting radiation loss, will be nearest to (a) 58ºC (b) 68ºC (c) 73ºC (d) 78ºC

Answer 1

Answer:

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Explanation:

Let the final temperature of the system be T.

Thus the heat absorbed by the cold water and beaker system=200×1×(T−20)+20×1×(T−20)

Heat lost be the hot water=440×1×(92−T)

From conservation of heat energy,

200×1×(T−20)+20×1×(T−20) =440×1×(92−T)

⟹T=68

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