posted by Amalia on .
Mercury, with a freezing point of -38.8°C, is the only metal that is liquid at room temperature. How much heat energy (in joules) must be released by mercury if 1.30 mL of the metal is cooled from room temperature (18.0°C) to -38.8°C and then frozen to a solid? (The density of mercury is 13.6 g/cm3. Its specific heat is 0.140 J/g · K and its heat of fusion is 11.4 J/g.)
Is this the type of question where you have to draw a heating curve and get your information after you draw the heating curve? What do I have to do after that?
Actually you don't need to draw a heating curve although I don't know that there is anything wrong with doing that. There are two equations that will solve any of these problems that go from one temp to another. ONe is for a phase change and the other is when there is no phase change.
For heat released going from 18 to -38.8, you have
q1 = mass x specific heat x (Tfinal-Tinitial). That will take you from 18 (room T) to the freezing point of Hg BUT IT ALWAYS A LIQUID IN THAT RANGE.
Then you want to freeze it. That formula is for a phase change.
q2 = mass x heat fusion
(Note: for the phase change from vapor to liquid, or the reverse, the formula is almost the same; i.e., q = mass x heat vaporization/condensation)
Then you just add the qs together to obtain the total q.