5 g of water at `30^(@)`C and 5 g of ice at `-20^(@)`C are mixed together in a caloritmeter. The water equivalent of calorimeter is negliglible and specific heat and latent heat of ice are 0.5 cal/`g^(@)`C and 80 cal/g repsectively. The final temperature of the mixture is

Options

A. `0^(@)`C
B. `-8^(@)`C
C. `-4^(@)`C
D. `2^(@)`C

Explanation

To solve the problem of mixing 5 g of water at 30°C with 5 g of ice at -20°C, we will calculate the heat exchanges involved and determine the final temperature of the mixture.

### Step-by-Step Solution:

1. **Calculate the heat required to raise the temperature of ice from -20°C to 0°C:**
- The specific heat of ice (s_ice) = 0.5 cal/g°C
- Mass of ice (m_ice) = 5 g
- Temperature change (ΔT_ice) = 0°C - (-20°C) = 20°C
...