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By Newton’s law of cooling, the rate of heat loss (cooling) is directly proportional to the difference in the temperatures between the body and its environment.

On the other hand, Specific heat capacity of a material is the change of its internal energy per unit mass for each unit change in its temperature.

In the case where two identical masses at the same initial temperature with different specific heat capacities are heated, the one with a lower specific heat capacity should gain temperature at a faster rate but both reaching the same equilibrium temperature eventually. Considering the equation Q=mc∆T, the internal energy gain by the mass with a lower c value is smaller. This information seems to bear much significance but in the following instance I could not fit them into the picture:

Now that these two masses are introduced to a much lower temperature, why then would the mass with a smaller heat capacity cool down faster, as asserted by my Physics textbook (with hardly any elaboration), if both masses have exactly the same temperature?

How then should I relate specific heat capacity of a material to its cooling rate, if there is a definitive way to relate the two at all?

Thank you.

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    $\begingroup$ You might wanna look at the derivation of Newton's Law of Cooling. What fundamental equation does it come from? Does the derivation include any approximations? If so, do the approximations hold in your "finite-difference in temperature" case? $\endgroup$
    – Agrim Arsh
    Commented Jun 14 at 5:06
  • $\begingroup$ Just to be clear, in your two identical masses example, are you talking about heat transfer by conduction or convection? $\endgroup$
    – Bob D
    Commented Jun 14 at 10:00
  • $\begingroup$ I think you need to differentiate between the rate at which energy is transferred to/from an object by means of heat, which depends on temperature difference, and the rate at which the temperature of an object changes as a result of that energy transfer, which depends on the specific heat capacity. $\endgroup$
    – Bob D
    Commented Jun 14 at 10:16

1 Answer 1

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The basic equation is $$\frac{dU}{dt}=\frac{dQ}{dt}$$where $$\frac{dU}{dt}=mC\frac{dT}{dt}$$and $$\frac{dQ}{dt}=-hA(T-T_{surr})$$with A representing the heat transfer area of the object and h representing the "heat transfer coefficient" between the object and the surrounding fluid.

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  • $\begingroup$ You're using $U$ for both internal energy and heat transfer coefficient? $\endgroup$
    – Bob D
    Commented Jun 14 at 15:25
  • $\begingroup$ @BobD Oops. I’ll change it. Thanks $\endgroup$
    – Chet Miller
    Commented Jun 14 at 15:30

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