elsie diaz asks:

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


(4) m = 12357.45 J/ 200 x 10^3 J/kg = 6.18 x 10^-2 kg 

of liquid nitrogen.

   *              *
   * m = 61.8 gm  *
   *              *

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