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    • Thread Starter

    Hi everyone
    I have a hard question.
    Here are two graphics:

    The power rating of the heater is 54 W.
    The heater is switched on and readings of the temperature of the block are taken at
    regular time intervals. When the block reaches a constant temperature, the heater is
    switched off and then further temperature readings are taken. The variation with time t
    of the temperature ? of the block is shown in Fig. 3.2.

    The question asks me

    After the heater has been switched off, the maximum rate of fall of temperature is 3.7 K per minute.

    Estimate the specific heat capacity of the aluminum cylinder.

    I know that in doing these types of questions, you are required to find out by how much temperature the aluminum raises the water or some other material. But this question doesn't have any water. Doesn't it supposed to ask how the temperature of the surrounding air increased?

    The answer in the mark scheme is:

    power input = maximum rate of heat loss
    power = m × c × ?? / ?t
    54 = 0.96 × c × 3.7 / 60

    But I do not understand.
    Why does power input = maximum rate of heat loss?
    How is that true?
    Is there any other way to approach that question, like the air heating up?

    Thank you for all your help!
    • Thread Starter

    OH OH OH, I think I got it.

    If we look at the graph, the gain in temp decreases to 0 (the graph flattens)
    this is because heat input = heat loss


    the input rate = max/initial/just after heater is shut off, rate of temperature decrease (from above)

    So we use this principle (of heat input = heat loss) to complete the problem.

    Am I making sense?
    Thanks for your help!

    When the block has reached it's maximum temperature, it is losing heat at the same rate it is receiving it from the heater. So it is losing heat at 54 joule per second. =54W
    When the heater is removed, the block starts to cool down because it is losing heat at 54 joule per second.
    This results in the temperature of the block falling by 3.7 deg per minute.
    So you know how much heat the block loses in one minute and you know it's change in temperature as well as it's mass.
    This in the information required to find the specific heat capacity as shown in the answer given.

    Yes your second post is correct.
    • Thread Starter

    Thank you SIR!

    I didn't notice that simple thing. 54 in and so losses 54 out at least initially, which is max heat loss.

    Thank you for your kind help Stonebridge !!! It means a ton!
    Thanks again Stonebridge
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