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

Answer:

#-2.99kJ# of heat has been liberated.

Explanation:

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

www.chem.purdue.edu

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#