Equal volumes of water and alcohol when put in similar calorimeters take `100 sec` and `74 sec` respecively to cool from `50^(@)C` to `40^(@)C`. The thermal capacity of each calorimeter in numerically equal to the volume of either liquid. The specific gravity of alcohol is `0.8` if the specific heat capacity of water is `1 cal//gm`, the specific heat capcity of alcohol will

Options

A. `0.7cal//g^(@)C`
B. `0.8cal//g^(@)C`
C. `1.6cal//g^(@)C`
D. `1.8cal//g^(@)C`

Explanation

To solve the problem, we need to find the specific heat capacity of alcohol based on the cooling times of equal volumes of water and alcohol in similar calorimeters. Here's a step-by-step solution:

### Step 1: Define Variables
Let:
- \( V \) = volume of water = volume of alcohol (in grams, since the specific gravity of water is 1)
- \( C_w \) = specific heat capacity of water = 1 cal/g°C
- \( C_a \) = specific heat capacity of alcohol (unknown)
- \( t_w \) = time taken for water to cool = 100 seconds
...