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Answer :
Approximately 17.3 kcal of energy must be removed from 0.175 kg of oxygen at 25°C to liquefy it at -193°C.
To determine the energy required to liquefy oxygen, we need to consider two steps: cooling the oxygen from 25°C to its boiling point and then condensing it at its boiling point.
Step 1: Cooling the oxygen from 25°C to its boiling point
The specific heat capacity formula is given by:
q = m * c * ΔT
Where:
q is the heat energy
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
Converting the given mass of oxygen to grams:
m = 0.175 kg * 1000 g/kg
m = 175 g
Calculating the heat energy required to cool the oxygen to its boiling point:
ΔT = boiling point - initial temperature
ΔT = -193°C - 25°C
ΔT = -218°C
q1 = m * c * ΔT
q1 = 175 g * 0.218 cal/g-°C * -218°C
q1 = 8378.9 cal
Converting calories to kilocalories:
q1 = 8378.9 cal * (1 kcal / 1000 cal)
q1 = 8.3789 kcal
Step 2: Condensing the oxygen at its boiling point
The heat of vaporization is given as 50.9 cal/g.
Calculating the heat energy required to condense the oxygen:
q2 = m * heat of vaporization
q2 = 175 g * 50.9 cal/g
q2 = 8917.5 cal
Converting calories to kilocalories:
q2 = 8917.5 cal * (1 kcal / 1000 cal)
q2 = 8.9175 kcal
Total energy required:
Total energy = q1 + q2
Total energy = 8.3789 kcal + 8.9175 kcal
Total energy ≈ 17.2964 kcal
Therefore, approximately 17.3 kcal of energy must be removed from 0.175 kg of oxygen at 25°C to liquefy it at -193°C.
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