What is the amount of heat liberated (in kJ) from 327 g of mercury when it cools from 77.8°C to 12.0°C?

1 Answer
Jul 2, 2016

-2.99kJ2.99kJ of heat has been liberated.

Explanation:

For this problem we have to use the specific heat capacity equation:

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Based on what you've given me, we have the mass of the sample (m), the specific heat (c), and the change in temperature DeltaT. We just need to find q.

DeltaT is -65.8^oC because we have to subtract the initial temperature from the final temperature.
(12^oC - 77.8^oC).

Since you want the amount of heat liberated to have units of kJ, you need to convert 0.139 J/(gxx^oC) into a value that has units of (kJ)/(gxx^oC)

We can do that by using this relationship:

1000J = 1kJ
Divide 0.139 by 1000 to obtain 0.000139 (kJ)/(gxx^oC)

Now, all of the variables have good units, so we just have to multiply all of the given values together to obtain Q (energy transferred).

Q = 327cancelgxx(0.000139kJ)/(cancelgxx^ocancelC)xx-65.8^ocancelC

Q = -2.99 kJ