a) Refrigerators and freezers remove heat from food by evaporating volatile liqu
ID: 557039 • Letter: A
Question
a) Refrigerators and freezers remove heat from food by evaporating volatile liquids such as CF4 and then recompressing the vapor back into a liquid. Calculate what mass of water at 0.0 deg C can be converted into ice at 0.0 deg C when 1.00 kg of CF4 evaporates. (delta Hvap = 136 kJ/kg for CF4)
b) Water in a leather flask or a porous clay pot can be cooled in hot climates by evaporating water from the surface of the container. Calculate what mass of water can be cooled from 35 deg C to 20 deg C via the evaporation of 60 g of water
Explanation / Answer
Ans. #A. Amount of heat absorbed during vaporization of 1 kg CF4 = dHvap x Mass of CF4
= (136 kJ/ kg) x 1.0 kg
= 136 kJ
# The heat is absorbed from liquid water by CF4.
So, total heat lost from water = -136 kJ. The –ve sign indicates that heat is lost from water.
# Now, heat lost by water is given by-
q = m C - equation 1
Where,
q = heat gained
m = mass
C = enthalpy of fusion of ice
Or,
136 kJ = m x (333.55 kJ/ kg)
Or, m = 136 kJ / (333.55 kJ/ mol)
Hence, m = 0.407734 kg
Hence, required mass of water converted into ice at 0.00C = 0.407734 kg
= 407.734 g
#B. Total heat absorbed during valorization of 60.0 g water = dHvap x mass of water
= (2257.0 J / g) x 60.0 g
= 135420 J
# The heat absorbed during vaporization is extracted from liquid water to cool it.
So,
Amount of heat lost by water during cooling = 135420 J
Now,
Heat lost by water given by-
q = m s dT - equation 1
Where,
q = heat lost
m = mass
s = specific heat
dT = Final temperature – Initial temperature
Or,
135420 J = m x (4.184 J g-10C-1) x (35.0 – 20.0)0C
Or, 135420 J = m x 62.76 J g-1
Or, m = 135420 J / (62.76 J g-1)
Hence, m = 2157.74 g
Hence, required mass of water = 2157.74 g = 2.158 kg
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