Question

How much water at (20°C) is required to cool a 5kg burning wood log from 300°C to 80°C ?

Answer 1

$$t_1=80^0$$

$$t_2=300^0$$

$$t_3=20^0$$

$$c_cm_c(t_2-t_1)=\lambda m_w+c_wm_w(t_1-t_2)$$

$$m_w=\frac {1.7\times 10^3\times 5 \times 220^0}{2.3\times 10^6+4.2\times 10^6\times 60^0}=0.732 kg$$