5400kJ of heat energy was lost when some amount of steam condensed to water for drinking purposes at 15º C. What is the quantity of water collected? [L\(_f ...

The steam gives out energy in two distinct stages, and both must be included:

1. Condensation at \(100\,^\circ\text{C}\). Changing state releases latent heat at constant temperature: \(Q_1 = mL\). The supplied value \(2.26\times10^{6}\,\text{J kg}^{-1}\) is the specific latent heat associated with the liquid-vapour change for water, which is the change happening here; note that the subscript printed with it conventionally denotes fusion, whose value for water is \(3.34\times10^{5}\,\text{J kg}^{-1}\), so use the number given and read it as the latent heat of vaporisation, since no melting or freezing occurs in this process.
2. Cooling of the condensed water from \(100\,^\circ\text{C}\) to \(15\,^\circ\text{C}\): \(Q_2 = mc\,\Delta\theta\) with \(\Delta\theta = 85\,\text{K}\).

The total energy released is therefore \[Q = m\left(L + c\,\Delta\theta\right).\] Evaluating the bracket first: \[L + c\Delta\theta = 2.26\times10^{6} + 4200 \times 85 = 2.26\times10^{6} + 3.57\times10^{5} = 2.617\times10^{6}\,\text{J kg}^{-1}.\] With \(Q = 5400\,\text{kJ} = 5.4\times10^{6}\,\text{J}\), \[m = \frac{5.4\times10^{6}}{2.617\times10^{6}} = 2.06\,\text{kg}.\] About \(2.06\,\text{kg}\) of water is collected.

Two errors account for the other figures. Using the latent heat alone gives \(5.4\times10^{6}/2.26\times10^{6} = 2.39\,\text{kg}\), because it ignores the cooling from \(100\,^\circ\text{C}\) to \(15\,^\circ\text{C}\); using the cooling term alone gives \(5.4\times10^{6}/(4200\times85) = 15.1\,\text{kg}\), because it ignores the far larger latent heat. Notice the scale of the two contributions: condensing \(1\,\text{kg}\) of steam releases roughly six times as much energy as cooling that same kilogram of boiling water down to room temperature, which is why steam scalds so severely. Always convert kilojoules to joules before dividing, and check that a latent-heat stage has no temperature change attached to it.