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A calorimeter of water equivalent 100 gr...

A calorimeter of water equivalent `100` grams contains `200` grams of water at `10^@ C`. A solid of mass `500` grams at `45^@ C` is added to the calorimeter. If equilibrium temperature is `25^@ C` then, the specific heat of the solid is `("in cal"//g -.^@ C)`.

A

`0.45`

B

`0.1`

C

`4.5`

D

`0.01`

Text Solution

Verified by Experts

The correct Answer is:
A
Heat lost by solid = Heat gained by calorimeter and water
`m_(s)S_(s)(45^@-25^@)=(m_(c)S_(c)+m_(w)S_(w))(25^@ -10^@)`.
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