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.)Select one:a.4320 calb.14,300 calc.9560 cald.2160 cal
Question
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.)Select one:a.4320 calb.14,300 calc.9560 cald.2160 cal
Solution
To solve this problem, we need to consider two steps:
- Cooling the oxygen from 22°C to -183°C.
- Liquefying the oxygen at -183°C.
Step 1: Cooling the oxygen We can calculate the heat removed in this step using the formula for heat transfer: q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature.
m = 150 g c = 0.218 cal/g·°C ΔT = -183°C - 22°C = -205°C
So, q1 = 150 g * 0.218 cal/g·°C * -205°C = -7027.5 cal
The negative sign indicates that heat is being removed.
Step 2: Liquefying the oxygen The heat removed in this step is given by the formula q = mL, where m is the mass and L is the heat of vaporization.
m = 150 g L = 50.9 cal/g
So, q2 = 150 g * 50.9 cal/g = 7635 cal
Again, the heat is being removed, so we take this as negative.
The total heat removed is the sum of the heat removed in these two steps: q_total = q1 + q2 = -7027.5 cal + -7635 cal = -14662.5 cal
So, the answer is approximately -14663 cal. However, this is not one of the options given. It's possible that there's a mistake in the problem or in the given options.
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