When a block of metal of specific heat `0.1 cal//g//^@C` and weighing 110 g is heated to `100^@C` and then quickly transferred to a calorimeter containing `200g` of a liquid at `10^@C`, the resulting temperature is `18^@C`. On repeating the experiment with 400 g of same liquid in the same calorimeter at same initial temperature, the resulting temperature is `14.5^@C`. find
a. Specific heat of the liquid.
b. The water equivalent of calorimeter.

Explanation

Let s be the specific heat of the liquid and W be the water equivalent of the calorimeter.
Heat lost by the block = heat gained by (liquid `+` colorimeter) for the first case:
`implies 110xx0.1xx(100-18)=200xxs`
`xx(18-10)+Wxx1xx(18-10)`
`implies1600s+8W=902` .. (i)
For the second case:
`implies110xx0.1xx(100-14.5)=400xxs`
`xx(14.5-10)+Wxx1xx(14.5-10)`
`implies1800s+4.5W=940.5` ..(ii)
on solving Eqs. (i) and (ii), we get `s=0.48cal//g//^@C` and `W=16.6g`