Correct Answer - D
Here,
Mass of water, `m_(w) = 100 g`
Mass of ice, `m_(i) = 100g`
Specific heat of water , `s_(w) = 1 cal g^(-1) .^(@)C^(-1)`
Latent heat of fusion of ice, `L_(fl) = 80 cal g^(-1)`
Let T be the final temperature of the mixture
Amount of heat lost by water
`m_(w)s_(w)(DeltaT)_(w) = 100 xx 1 xx (50 - T)`
Amount of heat gained by ice
`m_(i)L_(fl) + m_(i)s_(w) (DeltaT)_(i) = 10 xx 80 + 10 xx 1 xx (T - 0)`
According of heat lost by water
`m_(i)L_(fl) + m_(i)s_(w) (DeltaT)_(i) = 10 xx 80 + 10 xx 1 xx (T - 0)`
According to principle of calorimetry
Heat lost = Heat gained
`100 xx 1 xx (50 - T) = 100 xx 80 + 10 xx 1 xx (T - 0)`
`500 - 10T = 80 +T`
`11T = 420` or `T = 38.2 .^(@)C`