Question

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?

Answer 1

`-2.99kJ` of heat has been liberated.

Explanation:

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

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`