`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

Options

A. 20 g of water
B. 20 g of ice
C. 10 g ice and 10 g of water
D. 5 g ice and 15 g of water

Answer

C

Explanation

`Q_1=10xx1xx10=100cal`
`Q_2=10xx0.50[0-(-20)]+10xx80`
`=(100+800)cal=900cal`
As `Q_1ltQ_2`, so ice will not completely melt and final temperature`=0^@C`.
As heat given by water in coolingup to `0^@C` is only just sufficient to increase the temperature of the ice from `-20^@C` to `0^@C`, hence mixture in equilibrium will consist of 10 g of ice and 10 g of water, both at `0^@C`