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Q12.3(a):
Explain why (a) Two bodies at different temperatures $T_1$ and $T_2$ if brought in thermal contact do not necessarily settle to the mean temperature $(T_1 + T_2)/2$.

Solution :

Step 1: Theoretical Foundation of Heat Exchange

When two bodies of masses $m_1$ and $m_2$ at initial temperatures $T_1$ and $T_2$ (where $T_1 > T_2$) are brought into thermal contact, heat flows from the hotter body to the colder body until thermal equilibrium is reached at a common temperature $T_f$. [Per the Zeroth Law of Thermodynamics and the Principle of Calorimetry].

Step 2: Energy Conservation Equation

Assuming the system is isolated from the surroundings, the heat lost by the hotter body must equal the heat gained by the colder body:

$-Q_{lost} = Q_{gained}$

$m_1 c_1 (T_1 - T_f) = m_2 c_2 (T_f - T_2)$

Where:

  • $m_1, m_2$: Masses of the two bodies.
  • $c_1, c_2$: Specific heat capacities of the materials.
  • $T_f$: Final equilibrium temperature.

Step 3: Derivation of the Final Temperature

Expanding the equation to solve for $T_f$:

$m_1 c_1 T_1 - m_1 c_1 T_f = m_2 c_2 T_f - m_2 c_2 T_2$

$m_1 c_1 T_1 + m_2 c_2 T_2 = T_f (m_1 c_1 + m_2 c_2)$

$T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2}$

Step 4: Analysis of the "Mean Temperature" Condition

The equilibrium temperature $T_f$ will equal the arithmetic mean $\frac{T_1 + T_2}{2}$ if and only if the denominator coefficients are identical, i.e., $m_1 c_1 = m_2 c_2$.

If the bodies have different masses ($m_1 \neq m_2$) or different materials (different specific heat capacities $c_1 \neq c_2$), then $m_1 c_1 \neq m_2 c_2$. Consequently, the final temperature is a weighted average rather than a simple arithmetic mean.

Body 1 $m_1, c_1, T_1$ Heat flow Body 2 $m_2, c_2, T_2$

Step 5: Logical Conclusion

Since the thermal inertia (heat capacity, $C = mc$) of the two bodies is rarely equal, the temperature change per unit heat transfer differs for each body. Therefore, the equilibrium temperature is determined by the ratio of the products of mass and specific heat capacity, not by simple geometric midpoint logic.

Final Solution: The equilibrium temperature is a weighted average $T_f = \frac{m_1 c_1 T_1 + m_2 c_2 T_2}{m_1 c_1 + m_2 c_2}$. It only equals $(T_1 + T_2)/2$ if the heat capacities $m_1 c_1$ and $m_2 c_2$ are exactly equal, which is not a universal condition.


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