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

To solve the problem step by step, we will use the principles of calorimetry, which states that the heat lost by the hot object (the metal block) is equal to the heat gained by the cold object (the liquid in the calorimeter) plus the heat gained by the calorimeter itself.

### Given Data:
1. Specific heat of metal block, \( s_{\text{block}} = 0.1 \, \text{cal/g}^\circ C \)
2. Mass of metal block, \( m_{\text{block}} = 110 \, \text{g} \)
3. Initial temperature of metal block, \( T_{\text{initial, block}} = 100^\circ C \)
4. Mass of liquid in first case, \( m_{\text{liquid1}} = 200 \, \text{g} \)
5. Initial temperature of liquid, \( T_{\text{initial, liquid}} = 10^\circ C \)
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