Question 9.3: The molar heat capacity of liquid water is 75.3 J/mol K. If ......
The molar heat capacity of liquid water is 75.3 J/mol K. If 37.5 g of water is cooled from 42.0 to 7.0°C, what is q for the water?
Strategy The heat flow is proportional to the amount of water and the temperature change. As before, we can do the calculation with the amount of water expressed in either grams or moles. Molar heat capacity is given but the amount of water is provided in grams. So we will use the molar mass to convert the amount from grams to moles. (We could also leave the amount in grams and convert the molar heat capacity into the specific heat.) Care must also be taken in defining the change in temperature. Such changes are always the final state minus the initial state. Because the water is cooling here, ΔT will be negative.
q=n C_{\mathrm{p}}\Delta T
=37.5\ {\mathrm{g}}\times{\frac{1\;{\mathrm{mol}}}{18.0\;{\mathrm{g}}}}\times{\frac{75.3\;{\mathrm{J}}}{\mathrm{mol\ ^{\circ}\text C}}}\times-35.0^{\circ}{\mathrm{C}}
= –5.49 × 10³ J = –5.49 kJ
The negative value indicates that the system (in this case, the water) has lost energy to the surroundings. Notice that as long as we correctly express ΔT as T_{\mathrm{fi nal}} – T_{\text{initial}}, the correct sign for q will result automatically.
Check Your Understanding If 226 kJ of heat increases the temperature of 47.0 kg of copper by 12.5°C, what is the molar heat capacity of copper?