# 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. Commented Oct 31, 2012 at 17:02