Question

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 1

Answer

1161 kJ

Explanation

i) Heat required to raise the temperature of liquid H2O from 24℃ to 100℃ (q₁)

q₁ = m × ∆T × C_s

= 431g × (100° - 24℃ ) × 4.184J/g℃

= 137051J

ii) Heat required to vapourise liquid water at 100℃ (q₂)

q₂ = ∆H_vap × m

= 2257J/g × 431g

= 972767 J

iii) Heat required to raise the temperature of steam from 100℃ to 157℃ ( q₃)

q₃ = m × ∆T × C_s

= 431g × 57℃ × 2.078J/g℃

= 51050J

iv) Total heat required(q)

q= q₁ + q₂ + q₃

= 137051J + 972767J + 51050J

= 1160868J

= 1161kJ