A metal block of mass 200 g is heated to 100 °C and dropped into 150 g of water at 20.0 °C in a…

Step 1. The water. $Q_w$ $= 0.150\ \text{kg} \times 4186\ \text{J/(kg K)}$ $\times 8.6\ \text{K}$ $= 5400$ J.

Step 2. The calorimeter. $Q_c$ $= 0.050\ \text{kg} \times 385\ \text{J/(kg K)}$ $\times 8.6\ \text{K}$ $= 166$ J.

Step 3. The metal. The block loses $5400 + 166 = 5566$ J as it cools by $100 - 28.6 = 71.4$ K. So $c = \dfrac{Q}{m\,\Delta\theta}$ $= \dfrac{5566\ \text{J}}{0.200\ \text{kg} \times 71.4\ \text{K}}$ $= 390\ \text{J/(kg K)}$.