19 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice `=0.5calg^(-1) .^(@)C^(-1)` and latent heat of fusion of ice `=80calg^(-1)`

Options

A. `0^(@)C`
B. `-5^(@)C`
C. `5^(@)C`
D. `10^(@)C`

Answer

C

Explanation

Here, specific heat of ice, `c_(ice)`= 0.5 cal `g^(-1^(@))C^(@)^(-1)`
Specific heat of water, `c_(water)` = 1cal `g^(-1_(@))C^(-1)`
Latent heat of fusion of ice , `L_(ice)` = 80 cal `g^(-1)`
Here ice will absorb heat while hot water will release it.
Let T be the final temperature of the mixture.
Assuming water equivalent of calorimeter to be neglected.
Heat given by water, `Q_(1) = m_(water)c_(water)DeltaT`
`=19 xx 1xx(30-T) = 570-19T`................(i)
Heat absorbed by ice,
`Q_(2) = m_(ice) xx c_(ice) xx[0-(-20)]+m_(ice)xxl_(fice) + m_(ice) xx c_(water) xx (T-0)`
`5 xx 0.5 xx 20 + 5 xx 80 xx 5 xx 1 xx T=450 + 5T`.............(ii)
According to principle of calorimerty, `Q_(1) = Q_(2)`
i.e. 570 -19T = 450 + 5T `rArr T=120/24=5^(@)`C