A hot iron ball of mass 0.2 kg is added into 0.5 kg of water at `10^@`C. The resulting temperature is `30^@`C. Calculate the temperature of hot ball. Specific heat capacity of iron = 336 J `kg^(-1) K^(-1)` and specific heat capacity of water = `4.2 xx 10^3" "J" "kg^(-1) K^(-1)`.

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the hot iron ball will be equal to the heat gained by the water.

### Step 1: Write the heat transfer equations
The heat lost by the iron ball (Q_ball) is equal to the heat gained by the water (Q_water).

\[ Q_{\text{ball}} = Q_{\text{water}} \]
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