Question

At 1 atm, how much energy is required to heat 75.0 g of H2O(s) at –10.0 °C to H2O(g) at 137.0 °C?

Answer 1

At 1 atm, how much energy is required to heat 75.0 g of H2O(s) at –10.0 °C to H2O(g) at 137.0 °C?

This is a problem which involves phase changing and there are total 5 stages.

1...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released.
To bring the ice from -10°C to 0°C (Δ T = 10°C) = 75.0g x 2.1 J/g/°C x 16°C = 2520 J = 2.52 kJ.

2...mC = Q: mass x Latenr heat of melting.
To melt the ice at 0°C to water at 0°C. = 75g x 334 J/g = 25050 J = 25.05 kJ.

3...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released
To bring the water from 0°C to boiling at 100°C = 75g x 4.184 J/g/°C x 100°C = 31380 J = 31.38 kJ.

4...mC = Q: mass x Latenr heat of vaporisation.
To vaporise the water to steam at 100°C = 75g x 2,260 J/g = 169500 J = 169.5 kJ.

5...mcΔT = Q: m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released
To heat the steam from 100°C to 137°C = 75g x 2.01 J/g/°C x 37°C = 5578 J = 5.578 kJ

Total heat required = 2.52 kJ + 25.05 kJ + 31.38 kJ + 169.5 kJ + 5.578 kJ = 234.028 kJ.