1. **State the problem:** Calculate the specific heat capacity of Metal A using the method of mixtures. 2. **Given data:** - Mass of water, $m_w = 25$ g - Initial temperature of water, $T_{w_i} = 15^\circ$C - Final temperature of water, $T_{w_f} = 25^\circ$C - Mass of Metal A, $m_m = 20$ g - Initial temperature of Metal A, $T_{m_i} = 60^\circ$C - Final temperature of Metal A, $T_{m_f} = 25^\circ$C 3. **Formula used:** Heat lost by metal = Heat gained by water $$m_m c_m (T_{m_i} - T_{m_f}) = m_w c_w (T_{w_f} - T_{w_i})$$ where $c_m$ is the specific heat capacity of Metal A (unknown), and $c_w$ is the specific heat capacity of water, which is $4.18$ J/g/°C. 4. **Calculate temperature changes:** - $\Delta T_m = T_{m_i} - T_{m_f} = 60 - 25 = 35^\circ$C - $\Delta T_w = T_{w_f} - T_{w_i} = 25 - 15 = 10^\circ$C 5. **Substitute values into the equation:** $$20 \times c_m \times 35 = 25 \times 4.18 \times 10$$ 6. **Simplify the right side:** $$20 \times c_m \times 35 = 25 \times 41.8 = 1045$$ 7. **Isolate $c_m$:** $$c_m = \frac{1045}{20 \times 35}$$ 8. **Show cancellation:** $$c_m = \frac{1045}{\cancel{20} \times \cancel{35}} = \frac{1045}{700}$$ 9. **Calculate $c_m$:** $$c_m = 1.492857 \approx 1.49 \text{ J/g/}^\circ\text{C}$$ **Final answer:** The specific heat capacity of Metal A is approximately $1.49$ J/g/°C.