Question

A beaker contains 200gm of water. The heat capacity of the beaker is equal to that of 20gm of water. The initial temperature of water in the beaker is 20°C. If 440gm of hot water at 92°C is poured in it, the final temperature, neglecting radiation loss, will be nearest to?

Answer 1

Answer:

68∘ C

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∘ C