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Heat Transfer: Conversation of Energy problem.

  1. Sep 11, 2006 #1
    A 2 kg copper block, initially at 100°C, is brought in contact with a 2 kg plastic block, initially at 20°C. They are kept in contact long enough to reach the same equilibrium temperature. The specific heat of plastic is four times as much as the specific heat of copper.

    Which block will go through a larger temperature change? Why?

    Since this is a problem early in the first few chapters of a Heat Transfer class, I assume that the two blocks can be treated as a closed system and no heat is lost to the surroundings.

    Here is my working, which arrived at an answer that seems illogical.

    Energy lost by copper = Energy gained by plastic

    Since there is negligible change of other energies such as kinetic, potential, etc, assume that,

    Thermal energy lost by copper = Thermal energy gained by plastic

    [tex]mc \Delta T_{copper} = mc \Delta T_{plastic}[/tex]

    Since the masses of both blocks are equal,

    [tex](c \Delta T)_{copper} = (c \Delta T)_{plastic}[/tex]

    And since the specific heat of the plastic is four times that of the copper,

    [tex]c_{copper} \Delta T_{copper} = 4c_{copper} \Delta T_{plastic}[/tex]

    [tex]c_{copper}(T_{final} - 373) = 4c_{copper}(T_{final} - 293)[/tex] (temperatures in Kelvin)

    [tex]T_{final} - 373 = 4T_{final} - 1172[/tex]

    [tex]T_{final} = 266[/tex]

    Therefore, [tex] \Delta T_{copper} = 266 - 373 = -107 K[/tex]

    and [tex] \Delta T_{plastic} = 266 - 293 = -27 K[/tex]

    Since the cooler plastic block must gain heat from the warmer copper block, it is impossible that the temperature change for the plastic block could be negative. What's wrong with my calculations?

    Edit: Latex seems to format properly now, thanks Pseudo.
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 12, 2006 #2
    I think you mean \delta t. ;)
    Let me go back and see if I get what you did.
    The copper block is at 100C, and the plastic is at 20C... THUS, keeping that in mind, you're supposed to formulate your equation as follows:
    mc1 (Ti - Tf) = mc2 (Tf - Ti)
    WHY? Well, because obviously the copper block's going to cool down... but the plastic's going to heat up. You can't have one of them being positive and the other being negative and make them equal-- you have to adjust your heat to reflect that.
    In any case, you have...
    c2 = 4c1, Ti (for the left) = 100, Tf (for the right) = 20C, m = m so you get:
    mc1 (100 - Tf) = 4mc1(Tf - 20)
    Solve for Tf and compare the temperature differences. :)
  4. Sep 12, 2006 #3
    Oh, so it's backslash, thanks. :)

    I'm still not clear on why the delta t on the left is (Ti - Tf) and the delta t on the right is (Tf - Ti). Isn't delta t, the change in temperature, always the final temperature minus the initial temperature, regardless of whether the object in question is heating or cooling?

    I thought it worked like this: If an object were cooling, Tf - Ti would be negative, indicating a loss of temperature. If it were heating up, Tf - Ti would be positive, indicating a temperature increase. Could you please explain?
    Last edited: Sep 12, 2006
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