Question

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?

Answer 1

`85^@"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

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `m` is the mass of the sample
• `c` is the specific heat of the substance
• `DeltaT` is the change in temperature, calculated as the difference between the final temperature and the initial temperature of the substance

If you take `T_"f"` `""^@"C"` to be the final temperature inside the calorimeter, you can say that the heat gained by the piece that starts at `45^@"C"` is equal to

`q_"gained" = m * c * (T_"f" - 45)^@"C"`

Similarly, the heat lost by the piece that starts at `105^@"C"` will be equal to--keep in mind that this piece has twice the mass, or `2m`

`q_"lost" = 2m * c * (T_"f" - 105)^@"C"`

Use equation `color(darkorange)("( * )")` 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_"f" - 45 = -2T_"f" + 210`

Solve for `T_"f"` to find

`3T_"f" = 210 + 45`

`T_"f" = 255/3 = 85`

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

`color(darkgreen)(ul(color(black)("final temperature" = 85^@"C")))`

The answer is rounded to two sig figs.