# Chemistry; Calorimetry; Calculating Heat Capacity; Help Please?!

## One piece of copper jewelry at 105C has exactly twice the mass of another piece, which is at 45C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter?

Sep 8, 2017

${85}^{\circ} \text{C}$

#### Explanation:

The idea here is that the heat lost by the piece of jewelry that starts at a higher temperature will be equal to the heat gained by the piece of jewelry that starts at a lower temperature.

color(blue)(ul(color(black)(q_"gained" = - q_"lost")))" " " "color(darkorange)("( * )")

The minus sign is used here because the heat given off by the jewelry that cools off carries a negative sign!

Now, you can calculate the heat gained or given off by a substance by using the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, calculated as the difference between the final temperature and the initial temperature of the substance

If you take ${T}_{\text{f}}$ $\text{^@"C}$ to be the final temperature inside the calorimeter, you can say that the heat gained by the piece that starts at ${45}^{\circ} \text{C}$ is equal to

${q}_{\text{gained" = m * c * (T_"f" - 45)^@"C}}$

Similarly, the heat lost by the piece that starts at ${105}^{\circ} \text{C}$ will be equal to--keep in mind that this piece has twice the mass, or $2 m$

${q}_{\text{lost" = 2m * c * (T_"f" - 105)^@"C}}$

Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{( * )}}$ to say that--since both pieces of jewelry are made of copper, you know that they will have the same value for $c$

color(red)(cancel(color(black)(m))) * color(red)(cancel(color(black)(c))) * (T_"f" - 45) color(red)(cancel(color(black)(""^@"C"))) = - 2color(red)(cancel(color(black)(m))) * color(red)(cancel(color(black)(c))) * (T_"f" - 105)color(red)(cancel(color(black)(""^@"C")))

This is equivalent to

${T}_{\text{f" - 45 = -2T_"f}} + 210$

Solve for ${T}_{\text{f}}$ to find

$3 {T}_{\text{f}} = 210 + 45$

${T}_{\text{f}} = \frac{255}{3} = 85$

Therefore, you can say that the final temperature of the mixture is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{final temperature" = 85^@"C}}}}$

The answer is rounded to two sig figs.