31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams of water at 24.5 degrees Celsius. If the final temperature reached in the calorimeter is 32.5 degrees Celsius, what is the specific heat of the unknown substance?

Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g).

Question

31.5 grams of an unknown substance is heated to 102.4 degrees Celsius and then placed into a calorimeter containing 103.5 grams of water at 24.5 degrees Celsius. If the final temperature reached in the calorimeter is 32.5 degrees Celsius, what is the specific heat of the unknown substance?

Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g).

anonymous · Answer

I would use q = m x c x delta T right?

ciarán95 · Answer

Yes, you're exactly right, but this is where this heat energy transfer thing comes in again that you were wondering about earlier. The unknown substance is at a higher temperature than the water, so the substance will transfer heat energy, q, to the water which will absorb this energy. So the energy lost by the substance is equal to the energy gained by the water.

$$Qsubstance = Qwater$$
$$mc DeltaT (substance) = mcDeltaT (water)$$.

We are looking for c (substance), its specific heat capacity. We know the temperature changes and the masses of both the unknown substance and water, so we are good to go to try and solve this equation. Subbing in the values, we get:

$$(31.5g)(c)(102.4C - 32.5C) = (103.5g)(4.18 J/(C . g))(32.5C - 24.5C)$$

Its important that you make sure the delta Ts are kept positive (largest - smallest value) in both cases to get the correct answer. With a bit of manipulation, hopefully you should be able to solve for c (substance). Hope that helps! :)

anonymous · Answer

so far I got 
$$(31.5g)(c)(69.9c) = (103.5g)(4.18)(8c)$$

what do I do now?

ciarán95 · Answer

You need to solve for c (substance). To do this we need to get it on its own in the equation. If we divide both sides of the equation by 31.5 g, the 31.5 g on the left-hand side cancels and we are left with:
$$c(69.9) = \frac{ (103.5)(4.18)(8) }{ (31.5)}$$.

We can do the same thing by dividing both sides of the equation by 69.9 Celsius in order to cancel it from the left hand side and leave c on its own:
$$c = \frac{ (103.5)(4.18)(8) }{ (31.5)(69.9) }$$

This is the c (unknown substance) which we are looking for, and you can find the answer now using a calculator.

anonymous · Answer

I get it know, thanks! :)