Question

A 130.3 g piece of copper (specific heat 0.38 J/g･°C) is heated and then placed into 400.0 g of water initially at 20.7°C. The water increases in temperature to 22.2°C. What is the initial temperature of the copper? (The specific heat of water is 4.18 J/g･°C).

Answer 1

Answer:

Approximately $72.9\; \rm ^\circ C$, assuming that this system is insulated properly.

Explanation:

Calculate the amount of energy that the water here has gained (using the increase in its temperature.)

$\begin{aligned} Q(\text{water}) &= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T(\text{water}) \\ &=4.18 \times 400.0 \times (22.2 - 20.7) \approx 2508.0\; \rm J \end{aligned}$.

Assume that this system (copper and water) is well-insulated and has no heat exchange with its surroundings. The amount of energy that water gained should be equal to the of energy that copper has lost. However, the sign of the energy change of copper should be negative because copper had lost energy:

$Q(\text{copper}) = -Q(\text{water}) \approx -2508.0\; \rm J$.

Calculate the temperature change of this piece of copper. Note, that this temperature change is the difference between the initial temperature of the copper piece and the final temperature $22.2\; \rm ^\circ C$ of the system.

$\begin{aligned}\Delta T(\text{copper}) &= \frac{Q(\text{copper})}{c(\text{copper}) \cdot m(\text{copper})} \approx \frac{-2508.0}{130.3 \times 0.38} \approx 50.65\; \rm ^\circ C\end{aligned}$.

Calculate the initial temperature of this piece of copper given its final temperature and the change to its temperature:

$\begin{aligned}T(\text{copper, initial}) &= T(\text{copper, final}) - \Delta T(\text{copper}) \\ &\approx 22.2 + 50.65 \approx 72.9\; \rm ^\circ C\end{aligned}$.