Answer:
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
