How much heat would be released by the condensation of 5.40 g of steam at 100 °C and the subsequent cooling of the water to 25 °C?
[Δ_vapH = 40.7 kJ/mol at 100 °C;C_p for H₂O(l) is 4.18 J g^-1 °C^-1]

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat released during the condensation of 5.40 g of steam at 100 °C:

n(H2O) = 5.40 g / 18.02 g/mol = 0.299 mol

ΔH1 = n(H2O) x ΔHvap = 0.299 mol x 40.7 kJ/mol = 12.2 kJ

Next, we need to calculate the amount of heat released during the cooling of the water from 100 °C to 25 °C:

ΔT = 100 °C - 25 °C = 75 °C

q2 = m(H2O) x C(H2O) x ΔT = 5.40 g x 4.18 J/g°C x 75 °C = 1390.4 J = 1.3904 kJ

Therefore, the total amount of heat released is:

qtotal = ΔH1 + q2 = 12.2 kJ + 1.3904 kJ = 13.5904 kJ ≈ 13.9 kJ

So, the answer is choice C.

How Much Heat Would Be Released by the Condensation of 5.40