Question

Calculate the heat of fusion of ice according to the following laboratory data obtained by a not-too-precise student. Mass of calorimeter cup, 200 g, specific heat of the cup 0.11 , mass of water in the cup, 400 g, initial temperature of water 50^oC, mass of ice, 50 g, initial temperature of ice 0^oC, final temperature 30^oC

Answer 1

$$m_wc_w(T_w-T)=m_iL_f+m_wc_w(T-T_i)+m_cc_c(T-T_i),$$ $$L_f=\dfrac{m_wc_w(T_w-T)-m_wc_w(T-T_i)-m_cc_c(T-T_i)}{m_i},$$

L_f=\dfrac{400\ g\cdot1.0\ \dfrac{cal}{g\cdot \!^{\circ}C}\cdot20^{\circ}C-50\ g\cdot1.0\ \dfrac{cal}{g\cdot \!^{\circ}C}\cdot30^{\circ}C-200\ g\cdot0.11\ \dfrac{cal}{g\cdot \!^{\circ}C}\cdot30^{\circ}C}{50\ g},

$$L_f=116.8\ \dfrac{cal}{g}.$$