Suppose there are two objects $A$ & $B$ of same mass; the former has specific heat capacity $C_A$ and the later has $C_B$ such that $C_A\gt C_B.$ Let $A$ be at temperature $T$ while $B$ be at $T+\Delta T.$ So, in order to attain thermal equilibrium, they have to attain the same temperature $T+ \frac{\Delta T}{2}.$ That means there would be a change of $\frac{\Delta T}{2}$ in both bodies. For that, $A$ must gain heat energy $Q_A= m\;C_A\cdot \frac{\Delta T}{2}$ while $B$ has to loss heat energy $Q_B=m\;C_B\cdot \frac{\Delta T}{2}.$ Now, $Q_A-Q_B\gt 0$ since $C_A\gt C_B.$

Then in order to attain the equilibrium temperature, $A$ has to get $Q_A$ energy which it receives partly from $B$ viz. $Q_B$ but where does the additional heat energy $Q_A-Q_B$ come from? It can't come from $B$ as that would deviate it from achieving the thermal temperature; $B$ can lose only $Q_B$ in order to attain $T+\frac{\Delta T}{2}.$ But where does $A$ get the extra energy $Q_A- Q_B$ from?

I'm confused as I thought so-far that when two bodies achieve thermal equilibrium, they do it by exchanging heat energy between themselves - the cooler body receives heat energy fro the hotter body till the equilibrium is reached.

But here, it seems that $A$ in order to achieve thermal equilibrium, needs energy $Q_A$ which cannot be provided by $B$ since the energy $Q_B$ it releases is less than $Q_A.$ Where would then the extra energy $Q_A-Q_B$ come from?

Let's say, somehow $A$ gained that energy from the surroundings; however what would happen if they were isolated from the surroundings?

I don't know where I'm mistaking. Is it true that the hotter body itself provide the required energy to the cooler body to attain thermal equilibrium?

Please help.

  • $\begingroup$ According to your description and choice of symbols, $C_\mathrm A$ and $C_\mathrm B$ are heat capacities, not specific heat capacities. $\endgroup$
    – user7951
    Nov 7, 2015 at 17:20
  • $\begingroup$ $Q_A$ has to equal $Q_B$. Your mistake comes from the fact that at thermal equilibrium, the temperature of the two objects is not at $T+\Delta T/2$ if $C_A \neq C_B$. $\endgroup$
    – carbenoid
    Nov 7, 2015 at 17:26
  • $\begingroup$ @swenger: I was guessing this reason but didn't have any logic to explain the point. Could you elaborate it on why the thermal temperature shouldn't be $T+\frac{\Delta T}{2}?$ $\endgroup$
    – user5764
    Nov 7, 2015 at 18:11
  • $\begingroup$ @Loong: Sorry for that. $\endgroup$
    – user5764
    Nov 7, 2015 at 18:13

1 Answer 1


Say the initial temperature of $A$ is $T_1$ and the initial temperature of $B$ is $T_2$. To determine the temperature ($T_f$) of $A$ and $B$ at equilibrium, we start with the equation $q_A=-q_B$, where $q_A$ and $q_B$ are the heat gained by $A$ and $B$ respectively as they reach equilibrium. We have $q_A=mc_A\Delta T_A=mc_A(T_f-T_1)$ and $q_B=mc_B\Delta T_B=mc_B(T_f-T_2)$. Thus the equation becomes $$mc_A(T_f-T_1)=-mc_B(T_f-T_2)$$ From here, we can solve for $T_f$. We can also note that if $c_A=c_B$, $T_f=\frac{T_1+T_2}{2}$ , which in your example is $T+\frac{\Delta T}{2}$. Otherwise, $T_f$ will not be the midpoint between $T_1$ and $T_2$.

  • $\begingroup$ You've pre-assumed that $q_A=-q_B.$ Is this just? $\endgroup$
    – user5764
    Nov 7, 2015 at 20:09
  • $\begingroup$ Since the system is isolated, $q_A=-q_B$ because of conservation of energy $\endgroup$
    – carbenoid
    Nov 7, 2015 at 22:05

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