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:  


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