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:
- $C_{sp} \times \ce{H2O} = q(\mathrm{calorimeter})$
- $q(\ce{H2O})/m(\ce{H2O}) \times \Delta T = q(\mathrm{calorimeter})$
- $q(\ce{H2O}) = q(\mathrm{calorimeter})$
- 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.