How much energy is required to change a 20 g ice cube from ice at -40°C to steam at 100°C?

Question

How much energy is required to change a 20 g ice cube from ice at -40°C to steam at 100°C?

Answer 1

There are 4 separate energies that must be added together to get the total energy.
1. The heat necessary to get the water (ice) from -40C to 0C
2. The latent heat of fusion to change the water from a solid to a liquid at 0C (no temperature change just a phase change only)
3. The heat necessary to get the water from 0 C to 100C
4. The latent heat of vaporization to change the water from liquid form to gas form (steam) at 100 C (no temperature change, phase change only)

All these 4 numbers should be calculated based on a mass of 20 grams.

For water the latent heat of fusion is: 334 kJ/kg
and the latent heat of evaporation is: 2260 kJ/kg
Simply to heat the water in its solid or liquid form takes 4.18 kJ / kg C

The mass is 0.02 kg

1) 4.18 kJ / kgC * 0.02kg * 40C = 3.34 kJ
2) 334 kJ / kg * 0.02 kg = 6.68 kJ
3) 4.18 kJ / kgC * 0.02kg * 100C = 8.36 kJ
4) 2260 kJ / kg * 0.02 kg = 45.2 kJ

Add 'em all up

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Answer 2

to advance the temperature to 0°C, the nice and cozy temperature required is Q1 = mc?T the position m = fifty 5/1000 kg c = 1943 J/kg.°C the position we take the c fee at -20°C (see the ref lower than) ?T = 0 - (-40) = 40°C Q1 = 4274.6 J to transform 0.055kg of water at 0°C to ice at 0°C, a piece replace takes position so we use Q2 = mL the position L = 3.33 * 10^5 J/kg is the latent warmth of fusion for water Q2 = 18315 J to advance the temperature to at least one hundred°C from 0°C, the nice and cozy temperature required is Q3 = mc?T the position c = 4182 J/kg.°C for water at 50°C will be used (see the ref lower than) ?T = one hundred - 0 = one hundred°C So Q3 = 23001 J to transform water at one hundred°C to steam at one hundred°C, the nice and cozy temperature required is that q4 = mL the position L = 2279 kJ/kg is the latent warmth of evaporation q4 = 125345 J to advance the temperature to at least one hundred forty°C from vapour at one hundred°C, the nice and cozy temperature required is Q5 = mc?T the position c = 4248 J/kg°C for steam at 120°C (yet another ref provides c = certain warmth means water vapor at 120°C - a million.996 kJ/kgC on the boiling element. probable has to do with the rigidity which change into no longer specfied) and ?T = one hundred forty - one hundred = 40 Q5 = 9345.6 J (Q5' = 4391.2 J if we use the 2d fee for c) the nice and cozy temperature required is the sum Q Q = Q1 + Q2 + Q3 + q4 + Q5 (Q5') = 180281.2 J (175326.8 J) = a million.803 * 10^5 J (a million.753 * 10^5 J). wish this facilitates