# How to calculate the heat capacity of a calorimeter?

I need to find the heat lost of an unknown metal dropped into a calorimeter with $70~\mathrm{g}$ $\ce{H2O}$.
The initial temperature and final temperature of the $70~\mathrm{g}$ $\ce{H2O}$ and the calorimeter are $21~^\circ\mathrm{C}$ and $34~^\circ\mathrm{C}$. I already know that the heat gained by the water is $3807.44~\mathrm{J}$.

The metal's starting temperature and mass are $100~^\circ\mathrm{C}$ and $180.45~\mathrm{g}$, but that didn't help me much as I don't know the $C_{sp}$ of the unknown metal.

What I don't know is how to find the capacity of the calorimeter, any thoughts?

I already tried a number of solutions that didn't work:

1. $C_{sp} \times \ce{H2O} = q(\mathrm{calorimeter})$
2. $q(\ce{H2O})/m(\ce{H2O}) \times \Delta T = q(\mathrm{calorimeter})$
3. $q(\ce{H2O}) = q(\mathrm{calorimeter})$
4. I knew the unknown metal was one of three metals (lead, aluminum, copper), so I tried finding $q(\mathrm{metal})$ using the 3 heat specific heat capacities, but it didn't work out since the data provided were only approximations.
• Since all you were given is the temperature and mass of the metal, the temperatures and mass of the water, I would say that you ignore the calorimeter and use the figures to calculate the heat capacity of the metal (or whatever they are asking for).
– LDC3
Jan 22, 2015 at 5:08
• If the question isn't specifically about measuring anything about calorimeter, I suggest you ignore it. But if by any means the examiner (or the teacher) may insist on the calorimeter having any special effects on the outcome then the question gets kinda unanswerable. Jan 22, 2015 at 11:17

This is impossible to answer. Usually you have to assume that when no calorimeter heat capacity is given, then it negligible (i.e. you only use the heat capacity of the 70g $\ce{H_2O}$). You know the temperature drop of the metal and the energy increase of the water, combine both to obtain the heat capacity of the metal.
You know the heat capacity of the water given by: $$C=4.148\ \mathrm{\frac{J}{g\ K}}\ \times 70\ \mathrm g = 290.36\ \mathrm{J/K}$$ Since the water gained $3807.44\ \mathrm J$, you know that the metal loss $3807.44\ \mathrm J$. Using $Q = mc\Delta T$ you get: $$c = \frac{Q}{m\Delta T} = \frac{{-3807.44\ \mathrm J}}{180.45\ \mathrm g\times -66\ \mathrm K} = 0.320\ \mathrm{\frac{J}{g~K}}$$
Using the Dulong–Petit law, which states that the molar heat capacity of a metal is approximately $3R$, we get: $$\frac{3R}{c} = \frac{3 \times 8.314\ \mathrm{\frac{J}{mol~K}}}{0.320\ \mathrm{\frac{J}{g~K}}} = 78.01\ \mathrm{g/mol}$$ $M=78.01\ \mathrm{g/mol}$ is closest to copper so I would guess that is the answer.