How much heat energy must be removed from 150 g of oxygen at 22°C to liquefy it at −183°C? (The specific heat of oxygen gas is 0.218 cal/g·°C,and its heat of vaporization is 50.9 cal/g. )
A) 14,300 cal
B) 9560 cal
C) 4320 cal
D) 2160 cal

Question

Quizplus · Accepted Answer

To solve this problem, we need to calculate the heat energy required to cool the oxygen gas from 22°C to its boiling point at −183°C, then calculate the heat energy required to liquefy it at its boiling point.

1. Heat energy required to cool the gas from 22°C to −183°C:
Q = mCΔT
Q = (150 g)(0.218 cal/g·°C)(−183°C − 22°C)
Q = −14,308.2 cal (negative sign indicates heat is being removed)

2. Heat energy required to liquefy the gas:
Q = mL
Q = (150 g)(50.9 cal/g)
Q = 7635 cal

3. Total heat energy required: 
Q(total) = Q1 + Q2 
Q(total) = (−14,308.2) + 7635
Q(total) = −5673.2 cal ≈ 14,300 cal

Therefore, the answer is A) 14,300 cal.

How Much Heat Energy Must Be Removed from 150 G