CATAGORY: Mechanics QUESTION: A 30 g ice cube at its melting point is dropped into an insulated container of liquid nitrogen. How much nitrogen evaporates if it is at its boiling point of 77 K and has a latent heat of vaporization of 200 kJ/kg? Assume for simplicity that the specific heat of ice is a constant and is equal to its value near its melting point.

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Since the liquid nitrogen is at its boiling point the heat from the ice cube goes into evaporating liquid nitrogen. This continues until the ice cube temperature has dropped to the temperature of the liquid nitrogen. The difference in temperature times the heat capacity of ice (about 2100 J/kg C) times the mass of the ice gives the extracted energy. Note (1K = 1C) (1) energy from ice = (273.15K - 77K) 2100 J/kg C (30gm x 1kg/1000gm) = 196.15 C 2100J/kg C (.03 kg) = 12357.45 J The heat of vapoiization is l = 200kJ/kg = 200 x 10^3 J/kg so the vapoization energy H is given by (2) H = ml where m is the mass of liquid nitrogen vaporized. If the energy from the ice goes into H then we have (3) 12357.45 J = m 200 x 10^3 J/kg so (4) m = 12357.45 J/ 200 x 10^3 J/kg = 6.18 x 10^-2 kg of liquid nitrogen. **************** * * * m = 61.8 gm * * * ****************