Question

At 1 atm, how much energy is required to heat 87.0 g of H2O(s) at –16.0 °C to H2O(g) at 169.0 °C?

Answer 1

mcΔT = Q:

where

m = mass; c = specific heat, ΔT = temp. difference; Q = Heat absorbed or Released.

To heat the ice from -16°C to 0°C (Δ T = 16°C)

= 87g x 2.1 J/g/°C x 16°C

ΔT = 2923.2 J or 2.92 kJ.

2. mC = Q: mass x Latenr heat of melting.

To melt the ice at 0°C to water at 0°C

= 87g x 334 J/g

= 29058 J or 29.06 kJ.

again

3.mcΔT = Q:

where m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released

To heat the water from 0°C to boiling at 100°C so ΔT = 100 ºC

= 87g x 4.184 J/g/°C x 100°C

ΔT = 36400 J or 36.4 kJ.

mC = Q: mass x Latenr heat of vaporisation.

To vaporise the water to steam at 100°C

= 87g x 2,260 J/g = 196620 J or 196.62kJ

5. mcΔT = Q:

where m = mass; c = specific heat: ΔT = temp. difference; Q = Heat absorbed or Released

To heat the steam from 100°C to 169°C

= 87g x 2.01 J/g/°C x 69°C

ΔT = 12066.03 J or 12.06 kJ

Total heat required = 2.92 kJ + 29.06 kJ + 36.4 kJ + 196.62 kJ + 12.06 kJ = 277.06 kJ