Four cubes of ice at `-10^(@)C` each one gm is taken out from the refrigerator and are put in `150 gm` of water at `20^(@)C`. The temperature of water when thermal equilibrium is attained. Assuming that no heat is lost to the outside and water equivalent of container is `46 gm`. (Specific heat capacity of water `= 1 cal//gm-^(@)C`, Specific heat capacity of ice `=0.5 cal//gm-^(@)C`, Latent heat of fusion of ice `= 80 cal//gm-^(@)C`)

Options

A. `0^(@)C`
B. `-10^(@)C`
C. `17.9^(@)C`
D. None of these

Answer

C

Explanation

Heat gained by ice = Heat lost by water + Heat lost by container
Initail temperature of container =`20^(@)C`
`4xx1/2xx10+4xx80+4xx1x(T-0)=196xx1xx(20-T)`
`20+320 + 4T = 196xx20-196T`
`200T = 196xx20-304T = (3580)/(200)=17.9^(@)C`.