Calculations involving specific latent heat
The specific latent heat of water is:
- \(l_{f}=3\cdot34\,\times10^{5}\,J\,kg^{-1}\) for fusion (solid→liquid) or freezing (liquid→solid)
- \(l_{v}=22\cdot64\times10^{5}J\,kg^{-1}\) for vaporisation (liquid→gas) or condensation (gas→liquid)
The relationship used to determine the heat energy required to change the state of a substance is:
- \(E_{h} = ml\)
- \(E_{h}\) is the energy required to change the state ( \(J\))
- \(l\) is the specific latent heat of the substance ( obtained from the exam Data Sheet) (\(J\,kg^{-1}\))
- \(m\) is the mass of the substance (\(kg\))
Question
Calculate the energy required to change \(0.65kg\) of ice at \(0^{\circ}C \) into water at \(0^{\circ}C \)
ice →water is fusion, so \( l_{f}=3.34\,\times10^{5}\,J\,kg^{-1}\)
\(E_{h}=ml_{f}\)
\(E_{h}=0.65\,\times 3.34\,\times10^{5} \)
\(E_{h}=217100\,J\)
Heat energy required to melt \(0.65 kg\) water is \(217 100 J\).
The same relationship applies when the change of state is in the opposite direction and latent heat energy is removed from a substance ie gas→liquid or liquid→solid
Question
\(5. 65 \times 10^{6}J\) of heat energy is removed from a mass of steam at \(100^{\circ}C\) to produce water at \(100^{\circ}C\).
Calculate the mass of water produced.
steam→water is condensation so \(l_{v} =22\cdot6\,\times10^{5}\,J\,kg^{-1}\)
\(E_{h}=ml_{v}\)
\(5.65\,\times10^{6}=m\,\times22.6\,\times10^{5}\)
\(m=\frac{5.65 \times10^{6}}{22\.6\times10^{5}}\)
\(m= 2.5kg\)
So \(2.5 kg\) of water is produced
