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.
THANK YOU!!!!!!
Yes, this is the type of question where you can utilize a heating curve to analyze the heat energy involved in cooling and freezing mercury. However, in this specific question, you do not need to draw a heating curve since you are given the necessary information to calculate the heat energy.
To solve this problem, you can break it into two steps: (1) cooling the mercury to -38.8°C and (2) freezing it from -38.8°C to a solid.
Step 1: Cooling the mercury to -38.8°C
To find the amount of heat energy released during this step, you need to calculate the heat lost by the mercury using the specific heat capacity formula: q = m * c * ΔT, where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, convert the volume of mercury (1.30 mL) to grams using its density of 13.6 g/cm³:
mass = volume * density = 1.30 mL * 13.6 g/cm³ = 17.68 g
Next, calculate the change in temperature:
ΔT = final temperature - initial temperature = -38.8°C - 18.0°C = -56.8°C
Now, calculate the heat energy released during this step:
q1 = m * c * ΔT = 17.68 g * 0.140 J/g·K * (-56.8°C) = -142.46 J (the negative sign indicates heat release)
Step 2: Freezing the mercury to a solid
To find the amount of heat energy released during freezing, you need to multiply the mass of mercury by its heat of fusion.
Calculate the heat energy released during this step:
q2 = m * heat of fusion = 17.68 g * 11.4 J/g = 201.79 J
Finally, to find the total heat energy released, sum up the two steps:
Total heat energy = q1 + q2 = -142.46 J + 201.79 J = 59.33 J
Therefore, the amount of heat energy released by the mercury is 59.33 joules.