1. Given data

• Mass of unknown metal:
  $m_m = 192\,\text{g} = 0.192\,\text{kg}$
• Initial temperature of metal:
  $T_m = 100^\circ\text{C}$
• Mass of brass calorimeter:
  $m_b = 128\,\text{g} = 0.128\,\text{kg}$
• Specific heat of brass:
  $c_b = 394\,\text{J kg}^{-1}\text{K}^{-1}$
• Mass of water:
  $m_w = 240\,\text{g} = 0.240\,\text{kg}$
• Specific heat of water:
  $c_w = 4200\,\text{J kg}^{-1}\text{K}^{-1}$
• Initial temperature of water and calorimeter:
  $T_i = 8.4^\circ\text{C}$
• Final equilibrium temperature:
  $T_f = 21.5^\circ\text{C}$

2. Principle used

Heat lost by the hot metal = Heat gained by water + Heat gained by brass calorimeter.

So, $m_m c_m (T_m - T_f) = m_w c_w (T_f - T_i) + m_b c_b (T_f - T_i)$

where $c_m$ is the specific heat of the unknown metal.

3. Calculate temperature changes

• Metal cools through: $100 - 21.5 = 78.5^\circ\text{C}$
• Water and calorimeter warm through: $21.5 - 8.4 = 13.1^\circ\text{C}$

4. Heat gained by water

$Q_w = m_w c_w \Delta T$ $Q_w = 0.240 \times 4200 \times 13.1$ $Q_w = 13104\,\text{J}$

5. Heat gained by brass calorimeter

$Q_b = m_b c_b \Delta T$ $Q_b = 0.128 \times 394 \times 13.1$ $Q_b = 660.2752\,\text{J}$

6. Total heat gained

$Q_{\text{gain}} = Q_w + Q_b$ $Q_{\text{gain}} = 13104 + 660.2752$ $Q_{\text{gain}} = 13764.2752\,\text{J}$

7. Heat lost by metal

$Q_m = m_m c_m (78.5)$ $Q_m = 0.192\, c_m \times 78.5$ $Q_m = 15.072\, c_m$

By energy conservation, $15.072\, c_m = 13764.2752$

Thus, $c_m = \frac{13764.2752}{15.072}$ $c_m \approx 913.3\,\text{J kg}^{-1}\text{K}^{-1}$

8. Match with options

The closest option is:

$\boxed{916\,\text{J kg}^{-1}\text{K}^{-1}}$

So the correct option is D.

9. Option-wise check

• A: 458 — too small
• B: 1232 — too large
• C: 654 — too small
• D: 916 — matches calculated value

Therefore, the answer is: $\boxed{\text{D}}$