Let cube of mass 0.2 kg at `0^@C` be placed in a container whose temperature is `127^@C` . The specific heat of the container varies with temperature T as s =` p+qT^2` , where p = 120 cal/kg K and q = 0.03 `cal//kg K^3` . If the final temperature of the container is `27^@C` , what will be its mass? Take, latent heat of fusion of water `= 8 xx 10^4` cal/kg and specific heat of water = 1000 cal/kg K.

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the container will be equal to the heat gained by the ice cube.

### Step-by-Step Solution:

1. **Identify the heat gained by the ice cube:**
The ice cube at 0°C will first melt into water at 0°C and then the water will be heated to 27°C. The heat gained by the ice cube can be calculated using the formula:
\[
Q_{\text{ice}} = m_{\text{ice}} \cdot L_f + m_{\text{ice}} \cdot c_w \cdot \Delta T
...