# I need help with this question on Heat Capacity

A calorimeter has a Heat Capacity of $70 J/K$. There is $150g$ water with a temperature of $20^oC$ in this calorimeter. In this, you put a metal cube of $60g$ with a temperature of $100^oC$. The final temperature of all of this is $24^oC$.

What is the specific heat of this metal?

So, I'm having trouble understanding this problem, mainly the intuition behind this. I'm going to show you my progress ($cm$ = calorimeter, $w$=water, $m$=metal):

$Q_{cm} = 70 J/K$

$m_w = 150\times10^{-3} kg$

$\Delta T_w = 4^{\circ}C$

$m_m= 60\times 10^{-3} kg$

$\Delta T_m = 76$

I know I have to use the formulae:

• $Q = c\times m \times \Delta T$

• $Q = C \times \Delta T$

However, I'm lacking the insight needed the solve this problem. How would I have to go about solving the problem? I think I'm missing a fundamental insight needed to solve this problem. Can anybody help me with this?

In other words, I have the data, I have the correct formulas, but I don't know what to do with it!

• As I've gotten into some trouble with this in the past: I am not asking for an answer, I just want a push in the right direction. I don't know how to solve this problem, and I need help, not answer. I hope you people understand. – user14445 Oct 31 '12 at 17:02