Question

2. Calculate how much heat energy (J) is needed to change 150 g of water at 25°C into steam at 100°C. MUST show ALL steps of your work to receive credit.





(4 points)

Answer 1

Given: m = 150 g, Ti = 25.0 °C, Tboil = 100.0 °C.
Use cwater = 4.186 J·g⁻¹·°C⁻¹ and latent heat of vaporization Lv = 2260 J·g⁻¹.

1) Heat to raise liquid water from 25 °C to 100 °C:
ΔT = 100 − 25 = 75 °C
Q1 = m · c · ΔT = 150 g · 4.186 J·g⁻¹·°C⁻¹ · 75 °C
Q1 = 150 · 313.95 J = 47,092.5 J

2) Heat to vaporize water at 100 °C:
Q2 = m · Lv = 150 g · 2260 J·g⁻¹ = 339,000 J

3) Total heat required:
Qtotal = Q1 + Q2 = 47,092.5 J + 339,000 J = 386,092.5 J

Rounded appropriately: Qtotal ≈ 3.86 × 10^5 J (≈ 386 kJ).