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Homework Help: Specific heat capacity of copper

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid copper cylinder, 50 mm long and of 10 mm radius, is suspended in a vacuum calorimeter. Wound on the cylinder is a length of fine copper wire which is used as heater and resistance thermometer. Initially the resistance of the heater is 100.2 Ω. A current of 100 mA is then passed for 5 min, and then, when conditions are steady, the resistance of the heater is found to be 102.5 Ω
    Temperature coefficient of resistance of copper = [itex] a_1 = 4.1 \times 10^{-3} K^{-1}[/itex]
    Density of copper = [itex] \rho = 8.93 Mg m^{-3}[/itex]

    a) What is the specific heat capacity of copper?
    b)What assumptions are you making?
    c)What is the most important factor limiting the accuracy of the experiment?
    d)Why was it necessary to wait for conditions to become steady?

    2. Relevant equations
    [tex]Q = c \rho \pi r^2 l \Delta T[/tex]
    [tex]P = I R^2 = \frac{dQ}{ dt} [/tex]
    [tex]R_1=R_0 (1 + a_1 T)[/tex]

    3. The attempt at a solution
    For part a:
    [tex]R_1=R_0 (1 + a_1 T) <=> \left( \frac{R_1}{R_0} - 1 \right) \frac{1}{a_1} = T = 5.599 K[/tex]

    [tex]P = I R^2 = \frac{dQ}{ dt} = (R_0 (1 + a_1 T))^2 I = c \rho l \pi r^2 \frac{dT}{dt}≈ c \rho \pi r^2 l \frac{T}{\Delta t}[/tex]

    So I know [itex] \Delta T = 5.599 K [/itex] and
    [itex] \Delta t = 5 min = 300 s[/itex]

    The problem is that I don't know how either the resistance or temperature vary with time.
    I tried to calculate using the average value of resistance
    [tex]\bar{R} = \frac{R_0 + R_1}{2} = 101.35 Ω[/tex]
    [tex]P = I \left(\frac{R_0 + R_1}{2}\right)^2 = 1027,18 W[/tex]
    [tex] c = \frac{P}{\rho \pi r^2 l} \left(\frac{T}{\Delta t}\right)^{-1} = 392.6 J K^-1 kg^-1[/tex]

    The answer at the back of the book is c = 390 J K^-1 kg^-1 which a bit different from what I got.
    Am I missing something or is it just an error due to rounding off?

    For part b: I'm assuming that temperature varies linearly with time, and that resistance varies linearly with the temperature.Also I'm assuming that the heat transfered varies linearly with the temperature difference.

    For part c: I'd say the most important factor is not knowing exactly how temperature or resistance vary with time, hence having to use average values.

    For part d: I don't know why it has the conditions have to become steady. As it anything to do with the fact that once conditions become steady the resistance has a steady value and the transfer of energy rate is constant?
    Last edited: Sep 15, 2014
  2. jcsd
  3. Sep 15, 2014 #2


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    I do hope you used P = I2R, not IR2...:smile:

    You are worried about a 1% effect, which others might consider close to the experimental accuracy.

    I suppose you took r = 0.01 and l = 0.05 when calculating c ?
    Does that represent all the copper you heated up ?

    And I do find this a fairly theoretical exercise: a calorimeter also requires some heat to warm it up. And it loses heat to the environment.

    The exercise mentions "when conditions are steady" from which I could conclude that by then about 1W of heat flow leaks out to the environment (steady ##\equiv## dQ/dt = 0 **). Probably that's not what is meant, and they try to say something like "when the temperature rises at a constant rate"

    1W leakage means it's not such an expensive calorimeter... But one can always "assume" the calorimeter has zero heat capacity but still leaks so and so much.

    Problem is the leaked heat isn't used to increase the temperature of the copper...so then the calculation becomes quite a bit more complicated; probably not intended for this exercise.
  4. Sep 15, 2014 #3
    Ooops I did use P =I R^2, I'm so embarrassed right now.
    I don't know about what is meant by steady conditions. I thought it meant dQ/dt = 0, but why wouldn't that be what they are really trying to say?
    I don't know exactly how much copper is in the calorimeter, should I assume some length of copper is not taken into account ? Maybe the bits connecting the power source to the calorimeter ?
  5. Sep 15, 2014 #4


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    How about the heating wire itself ? With a wire of, say, .056 mm diameter you still need quite a length to get a 100 Ohm resistance... My guess is you end up with about 1% of the weight of the cylinder :smile:

    dQ/dt = 0 means you have to solve a differential equation (the heat leaking out is proportional to the temperature difference with the environment). But if the calorimeter is leaky, it probably also has some heat capacity of its own that cannot be ignored.
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