I answered this for you last night.
http://www.jiskha.com/display.cgi?id=1303519480
The freezing point of mercury is -38.8°C, is the only metal that is liquid at room temperature. What quantity of heat energy, in joules, must be released by mercury if 2.00 mL of mercury is cooled from 23.0°C to -38.8°C and then frozen to a solid? (The density of liquid mercury is 13.6 g/cm3. Its specific heat capacity is 0.140 J/g·K and its heat of fusion is 11.4 J/g.)
4 answers
Ahh sorry it was a mistake. Sorry :(
Wait but I tried out the question but I kept getting the wrong answer...
So if
[mass Hg x specific heat liquid Hg x (Tfinal-Tinitial) ] + (mass Hg x heat fusion)
(27.2 * 0.1410 * -61.8) + (27.2 * 11.4)
and I got the answer as -598 but the answer is wrong. What am I doing wrong?
So if
[mass Hg x specific heat liquid Hg x (Tfinal-Tinitial) ] + (mass Hg x heat fusion)
(27.2 * 0.1410 * -61.8) + (27.2 * 11.4)
and I got the answer as -598 but the answer is wrong. What am I doing wrong?
(27.2 * 0.1410 * -61.8) + (27.2 * 11.4)
First: You need a negative sign for the last term but you corrected for that because it's obvious you added the two terms as negative entities.
Second: If I go through your math I don't end up with -598 but something like 550 or so (approximately).
Third: The problem quotes 0.140 for specific heat liquid Hg and not 0.1410. That changes the final answer by approximately 2 J.
The problem, I think, is in the last two issues above. Let me know if this doesn't work.
First: You need a negative sign for the last term but you corrected for that because it's obvious you added the two terms as negative entities.
Second: If I go through your math I don't end up with -598 but something like 550 or so (approximately).
Third: The problem quotes 0.140 for specific heat liquid Hg and not 0.1410. That changes the final answer by approximately 2 J.
The problem, I think, is in the last two issues above. Let me know if this doesn't work.
2 answers