A vessel contains `110 g` of water. The heat capacity of the vessel is equal to `10 g` of water. The initial temperature of water in vessel is `10^(@)C`. If `220 g` of hot water at `70^(@)C` is poured in the vessel, the final temperature neglecting radiation loss, will be

Options

A. 70°C
B. 80°C
C. 60°C
D. 50°C

Answer

D

Explanation

Heat loss = heat gain
`rArr 220 xx 1 xx (70 - theta) = 110 xx 1 (theta - 10) + 10 (theta - 10)`
`rArr theta - 48.8^@ C ~~ 50^(@)C`.