I came across the following problem in my textbook today:

A container of negligible heat capacity contains $1~\rm{kg}$ of water. It is connected by a steel rod of length $10~\rm m$ and area of cross-section $10~\rm{cm}^2$ to a large steam chamber which is maintained at $100~^\circ\rm C\,.$ if initial temperature of water is $0~^\circ\rm C\,,$ find the time after which it becomes $50~^\circ \rm C\,.$ (Neglect heat capacity of steel rod and assume no loss of heat to surroundings) (specific heat of water $= 4180~\rm{J/kg}~^\circ C$)

The first step is to identify exactly what happens during the process and I suspect that I may be stuck in this very step itself. I know that when bodies having unequal temperatures are placed in thermal contact with each other, they will exchange heat energy via conduction (in this case) in an attempt to equalize their temperatures and be in thermal equilibrium with respect to the other body. Therefore, the temperature of both objects are changing simultaneously, yes? (As opposed to heat transfer via radiation in which Newton's law of cooling can be applied, and only one object changes its temperature; the temperature of the surroundings is assumed to be constant throughout the process.)

Given this, I was wondering what the fallowing equation (for the rate of change of heat transfer b/w two bodies held at different temperatures, placed in thermal contact w/ each other) even gives:

$$\frac Qt = \frac{KA(T_2-T_1)}{d}$$

If this rate itself is changing with time (until it attains steady state), then how to proceed with the problem? I tried this:

$Q=mC\Delta T$ and $Q/t=(\Delta T)KA/L$

$\implies Q/t=mC\Delta T/t=(\Delta T)KA/L$

Which gives me $mC/t=KA/L$

But clearly, this last equation is wrong, since it tells me that the time taken for heat transfer b/w two objects kept at any two temperatures is a constant given by rearranging:


What mistake have I made? Please note that I am not asking you to solve the problem for me; I have not understood how to use the concepts and formulae that I have learnt to solve this and request you to please give me a hint/nudge me in the correct direction, by also correcting the mistake that I have made above.

  • $\begingroup$ Have you studied any calculus? Newton's law of cooling is really a differential equation, $\frac{dQ}{dt}=...$. $\endgroup$
    – lemon
    Sep 13, 2016 at 8:03
  • $\begingroup$ Yes, I studied A.P calculus in high school(I have just graduated) and yes, you are right, but this question is a case of heat transfer by conduction, not radiation... $\endgroup$
    – user106570
    Sep 13, 2016 at 10:12
  • $\begingroup$ Who says that Newton's cooling only takes place when there is radiative heat transfer? $\endgroup$ Sep 13, 2016 at 12:11
  • $\begingroup$ Well, erm, my textbook :/ $\endgroup$
    – user106570
    Sep 13, 2016 at 13:49
  • $\begingroup$ Sorry, I am not convinced by your disingenuous claim that you are not asking us to solve the problem for you! You make a feeble attempt to start the problem without drawing a diagram or identifying any assumptions. Then after a few lines you stop calculating and ask what mistake you made - without getting any answer and without saying why you think you have gone wrong! Then you ask for a "nudge" in the right direction. Altogether this seems to me to be too lazy. You are getting into the bad habit of doing a little bit of thinking them asking someone else to sort it out for you. $\endgroup$ Sep 14, 2016 at 0:13

1 Answer 1


I suspect the question expects you to take the steam end of the rod to be fixed at 100°C. It uses the phrase large steam chamber suggesting it's large enough not to have any significant temperature change as it transfers heat to the water.

In that case the equation for the heat transferred to the water will be:

$$ \frac{dQ}{dt} = A (T_0 - T) \tag{1} $$

where $A$ is some constant that takes into account the conductivity and geometry of the connecting rod, $T_0$ is the (constant) temperature of the hot end and $T$ is the tempoerature of the water.

To make the equation more useful we note that the temperature of the water is related to the heat transferred by:

$$ T - T_i = \frac{Q}{C} $$

where $T_i$ is the initial temperature of the water and $C$ is the specific heat of the water. If we differentiate this we get:

$$ dT = \frac{dQ}{C} $$

And substituting this into equation (1) gives:

$$ \frac{dT}{dt} = \frac{A}{C} (T_0 - T) $$

which you solve by rearranging and integrating:

$$ \int\frac{dT}{T_0 - T} = \int\frac{A}{C} dt $$

Now, you as what happens then the the hot body has a finite heat capacity so it is cooling down while the water is heating up. We need to be a bit more careful about our notation. Let's call the temperature of the hot body $T_{H}$ and the temperature of the cold body $T_{C}$, and the initial temperatures $T_{H0}$ and $T_{C0}$. The specific heats of the hot and cold bodies will be $C_H$ and $C_C$.

Now our equation (1) becomes:

$$ \frac{dQ}{dt} = A (T_H - T_C) \tag{2} $$

where both $T_H$ and $T_C$ change as the heat flows. The values of $T_H$ and $T_C$ are given by:

$$\begin{align} T_H &= T_{H0} - \frac{Q}{C_H} \\ T_C &= T_{C0} + \frac{Q}{C_C} \end{align}$$

And the cunning trick is to subtract the second equation from the first to get:

$$ T_H - T_C = (T_{H0} - T_{C0}) - Q\left(\frac{1}{C_H} + \frac{1}{C_C}\right) $$

If you now substitute for $T_H - T_C$ in equation (2) you get:

$$ \frac{dQ}{dt} = A \left[(T_{H0} - T_{C0}) - Q\left(\frac{1}{C_H} + \frac{1}{C_C}\right)\right] $$

This looks a bit messy, but it's just the equation:

$$ \frac{dQ}{dt} = D(1 - E\,Q) $$

where $D$ and $E$ are constants containing all the messy stuff. Then the solution is just:

$$ \int\frac{dQ}{1 - E\,Q} = \int D\,dt $$

  • $\begingroup$ Whoa, thanks sir! Also, I was wondering what mistake I made when I got $mc/t=KA/L$ in my steps. I was looking to integrate at some point but bam! The ∆$T$ 's cancelled. $\endgroup$
    – user106570
    Sep 13, 2016 at 23:46

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