Heat related problem -- Mercury solidifying liberates some energy

[JennyLee1989]

Homework Statement

A 7 kg sample of mercury is completely solidified and liberates 81.06 kJ of energy.
What is the original temperature of the mercury? (The melting point of mercury is
234 K, the heat of fusion of mercury is 11.3 kJ/kg, and the specific heat of mercury is
140 J/kgÂK.) 4) A) 2.0 K B) 240 K C) 850 K D) 580 K

Relevant Equations

Q=mc(Tf-Ti)-mL+mc(Tf-Ti)

I have used the heat equation and am not able to get the initial heat. This is what I have so far:
81060 = 7 * 140 * (change in temp) - (7 * 11300) + 7 * 140 ( change in temp)
I found the change in temp to be 81.71K. But that does not help me with the initial temperature!

 

[haruspex]

The specific heat you are given is for liquid mercury. It is much higher for solid mercury. So the question ought to have stated that it is only just completely solidified.

Also, please explain the rationale for subtracting the latent heat of fusion.

 

[JennyLee1989]

The specific heat you are given is for liquid mercury. It is much higher for solid mercury. So the question ought to have stated that it is only just completely solidified.

Also, please explain the rationale for subtracting the latent heat of fusion.

The specific heat you are given is for liquid mercury. It is much higher for solid mercury. So the question ought to have stated that it is only just completely solidified.

Also, please explain the rationale for subtracting the latent heat of fusion.

The specific heat you are given is for liquid mercury. It is much higher for solid mercury. So the question ought to have stated that it is only just completely solidified.

Also, please explain the rationale for subtracting the latent heat of fusion.

The subtraction is because the heat is being removed. Isn’t addition used for when heat is added and subtraction for when heat is removed? Also, we are instructed to use the specific heat that is given. I have a feeling that this is just a bad question.

 

[haruspex]

The subtraction is because the heat is being removed

It is all removal of heat, first cooling the mercury to freezing point, then freezing it all.

we are instructed to use the specific heat that is given.

Which is how I deduce that no further heat is to be removed once it is all frozen. If further cooling were intended you would need to be given the specific heat of solid mercury as well as a final temperature.

 

[JennyLee1989]

It is all removal of heat, first cooling the mercury to freezing point, then freezing it all.

Which is how I deduce that no further heat is to be removed once it is all frozen. If further cooling were intended you would need to be given the specific heat of solid mercury as well as a final temperature.

I changed my equation to Q = mc(change in temp) - mL and I still get the wrong answer. Here's the work:
81060 = 7 * 140 (234 - Tinitial) - ( 7* 11300). [You're right, the whole equation is losing heat]
The answer for T initial came to be 70.57K which is incorrect. The correct answer is 240K which is so odd. Isn't this question impossible to solve without knowing the final temperature? I just tried using 234K as the final because of what you wrote that no further heat is to be removed once it is all frozen.

 

[Doc Al]

I changed my equation to Q = mc(change in temp) - mL and I still get the wrong answer. Here's the work:
81060 = 7 * 140 (234 - Tinitial) - ( 7* 11300). [You're right, the whole equation is losing heat]

Careful with signs!

The answer for T initial came to be 70.57K which is incorrect.

Which, of course, makes no sense. (Probably a sign error in one of your terms.)

The correct answer is 240K which is so odd. Isn't this question impossible to solve without knowing the final temperature? I just tried using 234K as the final because of what you wrote that no further heat is to be removed once it is all frozen.

As @haruspex has pointed out, you can assume that the final temp is just the freezing temp of mercury, 234 K.

 

[JennyLee1989]

Careful with signs!Which, of course, makes no sense. (Probably a sign error in one of your terms.)As @haruspex has pointed out, you can assume that the final temp is just the freezing temp of mercury, 234 K.

If you plug in the numbers, the answer, 240K, still does not come out. I tried changing the signs after reading your warning and the final temp came out to 232K.
Either the equation that I'm using is wrong or the answer key is wrong. I am using 234K as the final temperature.
Can you please try to solve this and tell me if you come up with 240K? This is driving me crazy.

 

[haruspex]

81060 = 7 * 140 (234 - Tinitial) - ( 7* 11300)

Still wrong.
You know the initial temperature must be more than 234K, so now you have a positive term on the left equalling a sum of negative terms on the right.

tried using 234K as the final

Good.

 

[Doc Al]

Can you please try to solve this and tell me if you come up with 240K? This is driving me crazy.

I did and, rounding off to 2 significant figures, I get 240K.

 

[Doc Al]

Either the equation that I'm using is wrong or the answer key is wrong.

If you're still messing up the signs, rewrite the equation this way:
Total heat released = Heat released during cooling + Heat released during freezing

Each term in that equation must be positive.

 

[JennyLee1989]

If you're still messing up the signs, rewrite the equation this way:
Total heat released = Heat released during cooling + Heat released during freezing

Each term in that equation must be positive.

I got it! Change in temperature is 2K. So 2K + 234K is 236K and if we use two sig figs, then it is 240K.
Is that what you got?

 

[Doc Al]

I got it! Change in temperature is 2K. So 2K + 234K is 236K and if we use two sig figs, then it is 240K.
Is that what you got?

You got it!

 

[mjc123]

The answer 240 K is ridiculous. (I'm not criticising those who have tried to answer the question, but the person who wrote it.) If the final temperature is given as 234 K, and the change in temperature is 2K, the initial temperature is 236 K. Why round it to 240? Why approximate 2 to 6? Why not round T_f to 230 K, and imply that a 2 K change in temperature is a 10 K change? If "7 kg" is taken as 1 s.f., why not round both temperatures to 200 K? This shows an unintelligent approach to the use of significant figures, especially when it comes to a small difference between two large quantities. It may make sense to apply 2 s.f. (or even only 1) to the calculated temperature change, but not to the two temperature values.

 

[Doc Al]

I agree completely, @mjc123.

 

[Vanadium 50]

The answer 240 K is ridiculous.

That's not the answer. The answer is "B". Are you disputing B is the best answer?

 

[Richard R Richard]

Hello, when heat is extracted from the mercury, it reduces its temperature, when the temperature reaches the melting temperature, if heat is extracted, the mercury begins to solidify.
It is interpreted from the statement that this heat extracted is just what is necessary to finish solidifying the entire mass of mercury. The energy balance would then be
Q = m_ {Hg} Ce_ {Hg} (T_i-T_f) + m_ {Hg} Cs_ {Hg}
81060 J = 7 kg\, 140 J / kg \, K (T_i-234 K) + 7 kg \,11300 J / kg
You just have to clear the Initial temperature T_i.
When a body loses heat its temperature drops.
To get to 234K, it is necessary that the mercury had been a higher temperature.
Edit: If you do the math, you will come up with an answer that is "not" offered as possible.