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Many friends in this forum told me that the rate of transfer of heat mainly depends on the difference in temperature rather than difference in specific heat capacity.

 

Your friends are right. When you study Thermodynamics, you study two things. Thermodynamics AND Heat Transfer. Thermodynamics and the concept of specific heat works without acknowledging the rate in which it happens. They usually assume enough time is given such that the system reaches equilibrium. Heat Transfer deals with the mechanism in which "heat" moves about.

 

In a Temperature-time graph, you cannot conclude information on the specific heat. You can tell from the graph however that liquid A freezes at a higher temperature than liquid B. And you can tell that liquid B dissipates heat at a faster rate than liquid A. Similarly I can also tell that A takes a longer time to solidify than B. I can't conclude much regarding its specific heat or heats of fusion however. Kind of like chemical kinetics if you've studied that. Rates of reaction not related to Change in Internal energy.

 

Liquid B may have had a high specific heat (high heat content), but it just gave away all that heat really really quickly. As for liquid A, it may have had a low specific heat (and thus low heat content), but it doesn't conduct its heat to air very well. A similar argument can be made for the freezing process.

 

To test the specific heat of liquids. You need to have the two samples at the same temperature then put them into a volume of water and measure the equilibrium temperature of the water. The one with more heat content will definitely have more heat to pass on to the surrounding water thus raising its temperature more. And again, similarly for heats of fusion, you can melt the two substances in water and see the equilibrium temperature of water.

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Many friends in this forum told me that the rate of transfer of heat mainly depends on the difference in temperature rather than difference in specific heat capacity.

here's an example' date=' what's the answer from you?[/quote']

 

The temperature change won't be linear like in the graph because the temperature difference will constanly decrease. (assuming the air temp doesn't change)

You will have an exponential curve. like unloading a capacitor with a resistor.

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In a Temperature-time graph, you cannot conclude information on the specific heat.

 

 

Sure you can, since they are equal mass and are exchanging ther heat with the same reservoir. The one with the larger initial slope has the smaller heat capacity - it transfers about the same amount of energy in some small time interval, since the temperature difference is the same, but its temperature drops more.

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Sure you can, since they are equal mass and are exchanging ther heat with the same reservoir. The one with the larger initial slope has the smaller heat capacity - it transfers about the same amount of energy in some small time interval, since the temperature difference is the same, but its temperature drops more.

 

You assume that their heat transfer is the same at temperature T for both.

If one has twice the surface in contact with the cooler body it will cool faster. (with the same heat capacity)

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Sure you can, since they are equal mass and are exchanging ther heat with the same reservoir. The one with the larger initial slope has the smaller heat capacity - it transfers about the same amount of energy in some small time interval, since the temperature difference is the same, but its temperature drops more.

But the rate of energy loss for them may not be necessarily the same, or the information in the graph is enough to compare their comparison?

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Sure you can, since they are equal mass and are exchanging ther heat with the same reservoir. The one with the larger initial slope has the smaller heat capacity - it transfers about the same amount of energy in some small time interval, since the temperature difference is the same, but its temperature drops more.

 

I would agree with that given that dQ/dt for both the materials are the same (they are dissipating the same amount of heat to the environment, or equally good at conducting heat to their surroundings), because dQ/dt is proportional to 1/mc x dT/dt. But there is no reason to assume so.

 

So through the reasoning in my earlier post, I'd like to challenge this statement. I'd like proof to see through the heat transfer equations that indeed

 

dT/dt not dependant on any other constants like thermal conductivity or diffusivity unless that can also be shown to be directly linked to a material's specific heat Cp, which would mean given a material's Cp we could determine its thermal conductivity and such.

 

P.S.considering the time before freezing to be fair. Because one liquid happened to have a higher freezing point than another doesn't conclude anything about its specific heat.

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You assume that their heat transfer is the same at temperature T for both.

If one has twice the surface in contact with the cooler body it will cool faster. (with the same heat capacity)

 

You are of course correct. I had assumed something not stated in the problem.

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But the rate of energy loss for them may not be necessarily the same, or the information in the graph is enough to compare their comparison?

 

I had made an assumption (see previous post) that was not actually given. Under that assumption the dQ would be the same, but it could also be different due to different surface areas.

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I have a similar problem. If I have a 100 gram copper sphere immersed in liquid nitrogen for an hour it would have a temperature of 77K. If I take it out an put it in "still" air at 300K, how long would it take it to come to 300K? Can that be calculated for "still" air?

 

The topic of heat transfer is a complicated one. Although it is often grouped together with a Thermodynamics course, the material itself can easily span a semester. You have to take into account the effect of not only conduction, but convection and radiation. Convection probably is what makes it most ugly.

 

Going back to your problem, with the information you have provided, it is not possible to calculate the time. The specific heat only deals with "equilibrium" situations. Also with regards to the problem originally posted here, unless the problem states one can assume dQa/dt = dQb/dt, no information regarding the specific heat nor conductivity can be deduced from the temperature-time graph. If your "solution" is otherwise, then the question has silently assumed the condition of equal heat transfer over time.

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I have a similar problem. If I have a 100 gram copper sphere immersed in liquid nitrogen for an hour it would have a temperature of 77K. If I take it out an put it in "still" air at 300K, how long would it take it to come to 300K? Can that be calculated for "still" air?

 

Can you measure the temperature of the sphere?

 

What you can do is making it half theoretical half imperial.

Put the sphere in the freezer (no liquid nitrogen required)

after a while measure the temp take it out wait 10minutes and measure the temp again.

The time and temp change will give you a constant that you need in the formula to calculate any temp change for any time.

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Is it true that if the copper sphere is at 77K and room temperature is 300K, it would rapidly rise in temperature by let's say 50K compared to the copper sphere being at 250K rising to 300K. In other words the rate of heat transfer depends on the difference big or small between the two temperatures?

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