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10 g of ice of 0^(@)C is mixed with ...

10 g of ice of `0^(@)C` is mixed with `100 g` of water at `50^(@)C` in a calorimeter. The final temperature of the mixture is [Specific heat of water `= 1 cal g^(-1).^(@)C^(-1)`, latent of fusion of ice `= 80 cal g^(-1)`]

A

`31.2^(@)C`

B

`32.8^(@)C`

C

`36.7^(@)C`

D

`38.2^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
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`
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