# 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?

Jul 2, 2016

$- 2.99 k J$ 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 $\Delta T$. We just need to find q.

$\Delta T$ is $- {65.8}^{o} C$ because we have to subtract the initial temperature from the final temperature.
(${12}^{o} C - {77.8}^{o} C$).

Since you want the amount of heat liberated to have units of kJ, you need to convert $0.139 \frac{J}{g {\times}^{o} C}$ into a value that has units of $\frac{k J}{g {\times}^{o} C}$

We can do that by using this relationship:

1000J = 1kJ
Divide 0.139 by 1000 to obtain $0.000139 \frac{k J}{g {\times}^{o} C}$

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).

$Q = 327 \cancel{g} \times \frac{0.000139 k J}{\cancel{g} {\times}^{o} \cancel{C}} \times - {65.8}^{o} \cancel{C}$

$Q = - 2.99 k J$