Question

A 46 g ice cube at its melting point is dropped into an insulated container of liquid nitrogen. Assume for simplicity that the specific heat of ice is a constant and is equal to 2100 J/kg⋅C∘

How much nitrogen evaporates if it is at its boiling point of 77 KK and has a latent heat of vaporization of 200 kJ/kgkJ/kg?

Answer 1

Given in the problem,

Mass of ice = 46 g = 0.046 kg

Ice Melts at 0°C

Nitrogen boils at 77 K.

Convert 77 K into deg C.

77K = 77 - 273 = -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= 2100 J/kg K

Suppose, mass of nitrogen evaporated = m kg.

Please note that, 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).

Now, we will have,

0.046 x 196 x 2100 = m x 200000

=> m = (0.046 x 196 x 2100) / 200000 = 0.0947 kg = 94.7 gm.

Therefore, mass of nitrogen evaporated = 94.7 gm (Answer)