1
$\begingroup$

Suppose we have two objects of masses $m_1$ and $m_2$, and temperature $T_1$ and $T_2$ respectively. Let them be in contact till they reach a final temperature $T$.

Question: $T$ will be closer to $T_1$ or $T_2$? Assume there is no heat exchange between the objects and surroundings. We can further assume the two objects are made of same material.

My intuition tells me that final temperature will be closer to object which has higher mass. But I have no idea whether my intuition is valid or not.

If someone can provide me a mathematical proof then it would be better.

$\endgroup$
2
$\begingroup$

Total energy: $E = cm_1T_1+cm_2T_2 = c(m_1+m_2)T = E_{after}$. It follows:

$T = \frac{m_1}{m_1+m_2}T_1 + \frac{m_2}{m_1+m_2}T_2$ and therefore, if $m_2 > m_1$ the term $\frac{m_2}{m_1+m_2}$ is larger than $\frac{m_1}{m_1+m_2}$ and hence, the temperature $T$ must be closer to $T_2$.

End result: $T$ is closer to temperature of Body with higher mass

$\endgroup$
  • $\begingroup$ You've implicitly assumed the heat capacities are the same. This is often a reasonable approximation, but it should definitely be noted, because even if the two objects are subject to all the same conditions except for mass and temperature, the heat capacity can depend explicitly on temperature. (For a toy example, if your system has a maximum energy then the heat capacity goes to zero as $T\to \infty$. You can also have quantum effects at low temperature.) If you let the heat capacities be $c_1,c_2$ then your answer basically goes through, but with "mass" replaced by "absolute heat capacity". $\endgroup$ – Ian Dec 19 '16 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.