Question

A 0.05 kg stone with a temperature of 75°C was placed in an isolated container filled with 0.05 kg of water at 15°C with a specific heat of 4180 J/(kg·°C). When The system reached thermal equilibrium, the temperature was 25°C. What is the specific heat of the stone?

Answer 1

$$m_sc_s(T_s-T)=m_wc_w(T-T_w)\\(0.05)(75-25)c_s=(0.05)(4200)(25-15)\\c_s=840\frac{J}{kg}$$