How much heat energy must be removed from 100 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. 13 700 calb. 9 560 calc. 4 320 cald. 2 160 cal

The right option is b. 9560 Cal

Explanation:

Q₁ + Q₂ = Total heat energy removed,

Where Q₁ = heat removed from as a result of latent heat of oxygen

Q₂ = Heat removed as a result of specific heat of oxygen

Q₁ = lm ... equation 1

Q₂ = cm (T₁-T₂) ... equation 2

l = heat of vaporization of oxygen, m = mass of oxygen, c = specific heat of oxygen. T₂ = Final temperature, T₁ = Initial temperature

Where, c = 0.218 cal/g⋅°C, l = 50.9 cal/g, m = 100 g, T₁ = 22°C, T₂ = - 183°C

Substituting this values into equation 1 and Equation 2 respectively,

Q₁ = 50.9 * 100 = 5090 cal

Q₂ = 0.218 * 100 * 22 - (-183)

Q₂ = 21.8 * 205 = 4469 cal

∴ Total Heat removed = Q₁ + Q₂

Total Heat removed = 5090 + 4469 = 9559 Cal.

Total Heat removed ≈ 9560 Cal

The right option is b. 9560 Cal