A 275-g sample of nickel at l00.0°C is placed in 100.0 g of water at 22.0°C. What is the final temperature of the water? Assume no heat transfer with the surroundings. The specific heat of nickel is 0.444 J/g·°C and the specific heat of water is 4.184 J/g·°C.

Question

Quizplus · Accepted Answer

To find the final temperature, we use the principle of conservation of energy, where the heat lost by nickel equals the heat gained by water. Let $T_f$ be the final temperature, $m_{Ni}$ and $m_{H_2O}$ be the masses of nickel and water, $C_{Ni}$ and $C_{H_2O}$ be their specific heats, and $T_{i,Ni}$ and $T_{i,H_2O}$ be their initial temperatures. The equation is:$$m_{Ni}C_{Ni}(T_f - T_{i,Ni}) = -m_{H_2O}C_{H_2O}(T_f - T_{i,H_2O})$$Plugging in the values:$$275 \times 0.444 \times (T_f - 100) = -100 \times 4.184 \times (T_f - 22)$$Solving for $T_f$ gives approximately 39.6°C.

A 275-G Sample of Nickel at L00