Latent Heat Calculations | CAIE 9702 Paper 4 Notes

Chapters

Paper 4

Physics9702

Latent Heat Calculations

Longer processes often contain several energy stages. Write one term for each stage, then use conservation of energy.

Build one energy equation for all stages

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Write one energy term for each stage that actually occurs. Then add the terms.

Mark the starting state and the final state of each material.
Write a specific-heat-capacity term for every temperature change and a latent-heat term for every change of state.
Set total energy lost equal to total energy gained when the system is thermally isolated.
Check the final state. Do not assume that all of a solid melts or all of a liquid vaporises.

Your turnquick check

Ice starts at 0 °C and finishes as water at 20 °C. Which two energy stages are needed?

Show worked answer

Q_{\mathrm{total}}=mL_f+mc_{\mathrm{water}}\Delta\theta

Separate a multi-stage calculation

Worked example

Ice added to warm water

37.0 g of ice at 0 °C is added to 208 g of water at 26.4 °C. The mixture is in a thermally insulated container whose heat capacity can be ignored. The final temperature is 10.3 °C. Find the specific latent heat of fusion of ice. The specific heat capacity of water is given below.

c_{\rm water}=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}

Calculate the energy lost by the warm water.
$\Delta\theta=26.4-10.3$
$\Delta\theta=16.1\ \mathrm{K}$
$Q_{\rm lost}=208\times4.18\times16.1$
$Q_{\rm lost}=1.40\times10^4\ \mathrm{J}$
The ice first melts. The melted ice then warms to 10.3 °C.
$Q_{\rm melt}=37.0L_f$
$mc=(37.0)(4.18)$
$mc=155\ \mathrm{J\,K^{-1}}$
$Q_{\rm warm}=155\times10.3$
$Q_{\rm warm}=1.59\times10^3\ \mathrm{J}$
Set energy lost equal to total energy gained and solve.
$37.0L_f=Q_{\rm lost}-Q_{\rm warm}$
$L_f=335\ \mathrm{J\,g^{-1}}$
Convert the unit to joules per kilogram.
$1\ \mathrm{J\,g^{-1}}=10^3\ \mathrm{J\,kg^{-1}}$
$L_f=3.35\times10^5\ \mathrm{J\,kg^{-1}}$

Answer

L_f=3.35\times10^5\ \mathrm{J\,kg^{-1}}

Based on (9702/41/M/J/25 Q3(c)).

The numerical value per kilogram is 1000 times the numerical value per gram because one kilogram contains 1000 grams.