Question

At 1 atm, how much energy is required to heat 35.0 g of H2O(s) at –20.0 °C to H2O(g) at 155.0 °C?

Answer 1

step-1

mcΔT = Q:

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

To heat the ice from -20 °C to 0°C (Δ T = 20 °C)

=35 g x 2.1 J/g/°C x 20°C

= 1435 J or 1.43 kJ.

mC = Q: mass x Latenr heat of melting

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

= 35 g x 334 J/g

= 11690 J or 11.69 kJ.

mcΔT = Q:

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

= 35g x 4.184 J/g/°C x 100°C

= 14644 J = 14.64 kJ.

mC = Q: mass x Latenr heat of vaporisation

To vaporise the water to steam at 100°C

= 35g x 2,260 J/g

= 79100 J = 79.100kJ.

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

To heat the steam from 100°C to 155°C

= 35g x 2.01 J/g/°C x 55°C

= 3869.25 J = 3.86 kJ

total heat required = add all together

= 1.43 kJ + 11.69 kJ + 14.64 kJ + 79.1 kJ + 3.86 kJ

= 110.72 kJ