Question 14.13: A 50.0-gram piece of copper metal at 80.0°C is placed in 100......
A 50.0-gram piece of copper metal at 80.0°C is placed in 100.0 mL (100.0 g) of water at 10.0°C. Assuming no energy is lost to the surroundings as heat, what will be the final temperature of the copperwater system? Take the molar heat capacities of Cu(s) and H_{2}O(l) as 24.4\ J·mol^{−1}·^{\circ }C^{−1} and 75.3\ J·mol^{−1}·^{\circ }C^{−1}, respectively.
When the hot copper metal is placed in the cool water, the copper will cool down and the water will warm up until they both come to the same final temperature. Let this (unknown) final temperature be T_{f}. The heat capacities of the copper and the water samples are
c_{P}[Cu(s)] = (50.0\ g)\left(\frac{1\ mol}{63.55\ g} \right) (24.4\ J·mol^{—1}·^{\circ}C^{–1}) = 19.2\ J·^{\circ}C^{–1}
c_{P}[H_{2}O(l)] = (100.0\ g)\left(\frac{1\ mol}{18.016\ g} \right) (75.3\ J·mol^{—1}·^{\circ}C^{–1}) = 418\ J·^{\circ}C^{–1}
Application of Equation 14.33, with Cu(s) as the high-temperature sample and H_{2}O(l) as the low-temperature sample, yields
c_{P_{h}}(T_{f} – T_{h}) + c_{P_{l}}(T_{f} – T_{l}) = 0 (14.33)
0 = c_{P_{h}}(T_{f} – T_{h}) + c_{P_{l}}(T_{f} – T_{l}) = (19.2\ J·^{\circ}C^{–1})(T_{f} – 80.0^{\circ}C) + (418\ J·^{\circ}C^{–1})(T_{f} – 10.0^{\circ}C)or, collecting like terms,
0 = [19.2\ J·^{\circ}C^{–1}+ 418\ J·^{\circ}C^{–1}]T_{f} – [(19.2)(80.0)J + (418)(10.0)J]from which we obtain
T_{f}=\frac{(19.2)(80.0)J + (418)(10.0)J}{19.2\ J·^{\circ}C^{–1} + 418\ J·^{\circ}C^{–1}} =\frac{5716\ J}{437\ J·^{\circ}C^{–1}}= 13.1\ ^{\circ}CThe temperature of the copper decreases by about 66.9°C (= 80.0°C – 13.1°C), whereas the temperature of the water increases by only 3.1°C (= 13.1°C – 10.0°C) because the heat capacity of the water is over twenty times greater than that of the copper.