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Heat transfer equation and internal energy

  1. Jul 6, 2013 #1
    The heat transfer equation is Rate = k•A•(T1 - T2)/d where k is the heat transfer coefficient of the material.

    But I'm feeling that the other material transferring the heat should be important too. I mean if I were to touch Styrofoam at 50 degrees and metal at 50 degrees, the metal feels hotter. But if i were to use the heat transfer equation, the coefficient would just be of my hand. So in both cases they would be the same? But we know that the metal feels hotter so doesn't the equation fail here? Or is it because for the heat transfer equation they must be of the same material? If so, for this case how do we calculate the heat transfer?

    Also, does internal energy affect the conductive ability? Say 2 substances with the same conductivity A and B. A has a higher specific heat capacity. So after heating it from the left side of the material for a while, comparing any point B will be hotter. So any atom would have more KE compared to A. So with more KE, the conductivity goes up?

    But the problem is that even though comparing each particle at the same distance from the heat source, the atom at B would have more KE. So that would mean that the particle on the right would also have more KE. In other words using the heat transfer equation, the temperature difference from one side of the material to the other might not be greater than for material A. So how can we tell if the conductivity of the material goes up?
  2. jcsd
  3. Jul 6, 2013 #2
    Heat trasfer coefficient is different for each material.
  4. Jul 6, 2013 #3


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    It feels hotter since metal can transfer a greater amount of heat to your hand in the time period of your touch than the styrofoam can. You hand have a greater increase in temperature, the feeling of being hot will reflect this.

    Do note that, syrofoam has as a large percentage of its volume a gas. The volumetric heat capacity ( http://en.wikipedia.org/wiki/Volumetric_heat_capacity ) of a gas is considered to be 1000 times less than that of a solid or liquid. A metal ball of 50 degrees in your hand will feel HOT, but the same size ball of styrofoam, with its less mass, will register much less.

    The k value is an indication of how 'fast' heat can flow through the material. For instance, a pipe made of metal ( with a high k value ) with hot liquid flowing through the pipe, can be considered to be at the same temperature of the liquid, both for the inside wall and outside wall of the pipe. Surround the pipe with adequate insulation such as styrofoam and the inside wall of the insulation will be at the temperature of the pipe but the outside wall of insulation will be nearly the temperature of the air.
  5. Jul 6, 2013 #4
    You are talking about a the transient heat transfer problem that applies when two materials with different temperatures and thermal properties are brought into contact with one another. Imagine a semi-infinite slab of material A that is brought into thermal contact with a semi-infinite slab of material B at time = 0. Materials A and B have differing thermal properties and densities: Cp, k, and ρ, where Cp is the heat capacity, k is the thermal conductivity, and ρ is the density. The initial temperatures of the two slabs at time t = 0 are uniform, and equal to TA and TB, respectively. What are the differential equations and boundary conditions that describe the thermal response of this system? Assume that material A occupies the region from -∞ < x ≤ 0, and material B occupies the region from 0 ≤ x < ∞. The differential equations for this system are as follows:

    [itex]\frac{\partial T}{\partial t}=\kappa_A \frac{\partial^2 T}{\partial x^2}[/itex] for -∞ < x ≤ 0

    [itex]\frac{\partial T}{\partial t}=\kappa_B \frac{\partial^2 T}{\partial x^2}[/itex] for 0 ≤ x < ∞

    where κA and κB are the respective thermal diffusivities:
    Initial conditions are:

    T = TA for -∞ < x ≤ 0 at t = 0

    T = TB for 0 ≤ x < ∞ at t = 0

    Note that, prior to bringing the two slabs together, there is a jump difference in temperature at x = 0. However, as soon as the slabs make contact, the temperature at the interface assumes a value (for all time) between T = TA and T = TB.

    The boundary condition at x = 0 is: [itex]k_A\left(\frac{\partial T}{\partial x}\right)^-=k_B\left(\frac{\partial T}{\partial x}\right)^+[/itex]
    where the - and the + apply to the gradients on each of the two sides of the interface. Far from the interface, the boundary conditions are:

    T → TA at x → -∞

    T → TB at x → +∞

    This completes the statement of the transient heat transfer problem. The solution to these equations can tell you what the temperature at the interface will be for all time, as well as the distribution of temperature as a function of time. But you can see from these equations precisely which material properties will play a role in determining the heat transfer behavior.
  6. Jul 8, 2013 #5
    Wow this is amazing! :) with this I would I be able to tell the temperature difference from one end of the slab and the other? So I can tell if the rate of heat flow increases or decreases too?
  7. Jul 8, 2013 #6
    Yes. You would be able to tell everything you want. Of course, if you are interested in what is happening from one end of the slab to the other, you would have to solve the problem for slabs of finite thickness. In the case of the heat transfer rate between the slabs, this would always decrease with time. For solutions to these types of problems, see Carslaw and Jaeger, Conduction of Heat in Solids (the bible), or, Transport Phenomena by Bird, Stewart, and Lightfoot.

  8. Jul 9, 2013 #7
    This is great! I don't understand the math behind it but knowing that there's a way to calculate the rates of the transfers intrigues me. When i'm proficient with the math I'll try it out haha!

    Thanks Chet!
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