How much nitrogen evaporates if it is at its boiling point of 77 K and has a lat

Question

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.

Explanation / Answer

Mass of ice = 27g => .027kg Ice Melts at 0°C Nitrogen boils at 77K. => 77K is -196°C Nitrogen has a latent heat of vaporization of 200 kJ/kg =>200000 J/kg Temperature change = 0°C-(-196°C)=196°C Specific heat capacity of ice= 2060 J/kg K Let 'm' in the equation, represent the unknown mass of nitrogen evaporated. Note: Since the ice at 0°C is placed into nitrogen at -196°C it wont melt as the temperature is now lower than its freezing point (0°C). .027 x 196 x 2060 = m x 200000 12516 = 200000 x m m = 12516/200000 m = .0544 kg

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