A 0.25 kg piece of metal at 120 °C is dropped into 0.60 kg of water at 18 °C in a 0.10 kg copper…

Step 1. Heat gained. $(0.60\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} + 0.10\ \text{kg} \times 385\ \text{J kg}^{-1}\text{K}^{-1}) \times 3.6\ \text{K} = 9180\ \text{J}$.

Step 2. Specific heat. $\dfrac{9180\ \text{J}}{0.25\ \text{kg} \times 98.4\ \text{K}} = 373\ \text{J kg}^{-1}\text{K}^{-1}$.