91Ó°ÊÓ

How much heat is given up when \(20 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed and cooled to \(20{ }^{\circ} \mathrm{C}\) ? \(\begin{aligned} \text { Heat change } &=\text { Condensation heat change })+(\text { Heat change of water during cooling) }\\\ &=m L_{v}+\mathrm{cm} \Delta T \\ &=(20 \mathrm{~g})(-540 \mathrm{cal} / \mathrm{g})+\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(20 \mathrm{~g})\left(20^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}\right) \\\ &=-12400 \mathrm{cal}=-12 \mathrm{kcal}=-50 \mathrm{~kJ} \end{aligned}\)

Suppose a 60 -kg person consumes 2500 Cal of food in one day. If the entire heat equivalent of this food were retained by the person's body, how large a temperature change would it cause? (For the body, \(\left.c=0.83 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} .\right)\) Remember that \(1 \mathrm{Cal}=1 \mathrm{kcal}=1000 \mathrm{cal}\). The equivalent amount of heat added to the body in one day is $$ \Delta Q=(2500 \text { Cal })(1000 \mathrm{cal} / \mathrm{Cal})=2.5 \times 10^{6} \mathrm{cal} $$ Then, by use of \(\Delta Q=m c \Delta T\), $$ \Delta T=\frac{\Delta Q}{m c}=\frac{2.5 \times 10^{6} \mathrm{cal}}{\left(60 \times 10^{3} \mathrm{~g}\right)\left(0.83 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)}=50^{\circ} \mathrm{C} $$

How much water vapor exists in a \(105-\mathrm{m}^{3}\) room on a day when the relative humidity in the room is 32 percent and the room temperature is \(20^{\circ} \mathrm{C}\) ? Saturated air at \(20^{\circ} \mathrm{C}\) contains \(17.12 \mathrm{~g} / \mathrm{m}^{3}\) of water.