How much heat (in kJ) is required to convert 431 g of liquid H2O at 24.0°C into steam at 157°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C.
Answer
1161 kJ
Explanation
i) Heat required to raise the temperature of liquid H2O from 24℃ to 100℃ (q1)
q1 = m × ∆T × Cs
= 431g × (100° - 24℃ ) × 4.184J/g℃
= 137051J
ii) Heat required to vapourise liquid water at 100℃ (q2)
q2 = ∆Hvap × m
= 2257J/g × 431g
= 972767 J
iii) Heat required to raise the temperature of steam from 100℃ to 157℃ ( q3)
q3 = m × ∆T × Cs
= 431g × 57℃ × 2.078J/g℃
= 51050J
iv) Total heat required(q)
q= q1 + q2 + q3
= 137051J + 972767J + 51050J
= 1160868J
= 1161kJ
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