Question

A 1-kg block of copper at 20°C is dropped into a large vessel of liquid nitrogen at 77K. How many kilograms of nitrogen boil away by the time the copper reaches 77K? (The specific heat of copper is 0.092 cal/g·°C. The heat of vaporization of nitrogen is 48 cal/g.)

Answer 1

Given data For copper: initial temperature of copper T_c_i = 20 degree C = (20 + 273) = 293 degree k final temperature of copper = 77 degree k specific heat capacity of copper = 0.092 cal/g degree c = 0.092 times 4.187/10^-3 = 385.2 J/kg degree k Mass of copper = 1 kg For Nitrogen: mass of Nitrogen M_N = ? Nitrogen temperature = 77 degree K (missing information = mention that T = 77 degree K is boiling point of Nitrogen). Latent that at vaporization L_V = 78 cal/g = 48 times 4.187/10^-3 = 200976 J/kg heat lost from copper = heat gain by Nitrogen m_c C_c Delta T = m_N l_V therefore m_N = m_c C_c Delta T/l_V = 1 times 385.2 times (293 - 77)/200976 m_N = 83203.2/200976 = 0.414 kg