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.)
I answered this for you last night.
http://www.jiskha.com/display.cgi?id=1303519480
Ahh sorry it was a mistake. Sorry :(
(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.
To find the quantity of heat energy released by mercury when it is cooled and frozen, we need to calculate two separate quantities of heat energy: one for cooling the liquid mercury and another for freezing it.
1. Heat energy for cooling (ΔQ1):
First, we need to calculate the mass of mercury. We are given the volume of mercury as 2.00 mL and the density of liquid mercury as 13.6 g/cm³. Using the formula:
mass = volume × density
mass = 2.00 mL × 13.6 g/cm³
mass = 27.2 g
Next, we need to find the temperature change. The initial temperature of mercury is 23.0°C, and the final temperature is -38.8°C. Therefore, the temperature change (ΔT1) is:
ΔT1 = final temperature - initial temperature
ΔT1 = -38.8°C - 23.0°C
ΔT1 = -61.8°C
Now, we can calculate the heat energy for cooling using the formula:
ΔQ = mass × specific heat capacity × temperature change
ΔQ1 = 27.2 g × 0.140 J/g·K × -61.8°C (Note: Celsius and Kelvin scales have the same unit difference)
ΔQ1 = -110.57 J (rounded to two decimal places)
2. Heat energy for freezing (ΔQ2):
To calculate the quantity of heat energy released during freezing, we need the mass of mercury (already calculated) and the heat of fusion.
The heat energy for freezing is given by the formula:
ΔQ2 = mass × heat of fusion
ΔQ2 = 27.2 g × 11.4 J/g
ΔQ2 = 310.08 J
To find the total heat energy released, we need to sum up the heat energy for cooling (ΔQ1) and the heat energy for freezing (ΔQ2):
Total heat energy released = ΔQ1 + ΔQ2
Total heat energy released = -110.57 J + 310.08 J
Total heat energy released = 199.51 J (rounded to two decimal places)
Therefore, the quantity of heat energy released by mercury when 2.00 mL of it is cooled from 23.0°C to -38.8°C and then frozen to a solid is approximately 199.51 Joules.
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?