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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 $

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

$\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!

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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

1 Answer 1

up vote 2 down vote accepted

Here's a hint: the heat released by the block of metal has to go someplace. Some of it will go into heating the water and some of it will go into heating the calorimeter. Write an equation linking these three heats together. Then write an equation using temperature differences and heat capacities instead of the heats.

Now you know the heat capacities involved with two of the heats. You can solve for the third one.

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