IPhO 1967 · Theory problem 3

Equal heat supplied to a hanging ball and a ball resting on a plane

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Consider two identical homogeneous balls, A and B, with the same initial temperatures. One of them is at rest on a horizontal plane, while the second one hangs on a thread (Fig. 6). The same quantities of heat have been supplied to both balls. Are the final temperatures of the balls the same or not? Justify your answer. (All kinds of heat losses are negligible.)

Figure 1

Fig. 6

Official English text of the problem.

Answers

  • Problem 3: In the usual elementary model, assuming positive thermal expansion, the hanging ball A becomes hotter than the resting ball B: \[ \boxed{T_A>T_B.} \]

Solution

Assume that the plane is rigid, the thread has a fixed length, and both balls expand uniformly when heated. We use the elementary approximation that their thermal-energy changes have the same dependence on temperature, neglecting the influence of supporting stresses.

Let each ball initially have radius \(R_0\) and temperature \(T_0\). Denote their common mass by \(m\) and the heat supplied to each by \(Q\).

When the radius increases by \(\Delta R\):

  • For A, the top attachment point stays fixed, so its centre of mass descends by \(\Delta R\).
  • For B, its lowest point remains on the plane, so its centre of mass rises by \(\Delta R\).

Their gravitational potential-energy changes are therefore \[ \Delta E_{g,A}=-mg\Delta R_A, \qquad \Delta E_{g,B}=+mg\Delta R_B. \]

The stationary supports do no work. Conservation of energy gives \[ Q=\Delta U+\Delta E_g, \] where common work against atmospheric pressure may either be neglected or included in the thermal-energy term. Thus \[ \Delta U_A=Q+mg\Delta R_A, \qquad \Delta U_B=Q-mg\Delta R_B. \]

For positive thermal expansion, both radius changes are positive. Consequently, \[ \Delta U_A>Q>\Delta U_B. \] With the assumed common, increasing relationship between thermal energy and temperature, this implies \[ \boxed{T_A>T_B.} \]

Explicitly, let \(c\) be the common specific heat capacity and \(\alpha>0\) the linear expansion coefficient, treated as constant over a small temperature interval. Writing \(\Delta T_i=T_i-T_0\), \[ \Delta U_i=mc\Delta T_i, \qquad \Delta R_i=\alpha R_0\Delta T_i. \] The energy balances become \[ Q=m(c-g\alpha R_0)\Delta T_A, \qquad Q=m(c+g\alpha R_0)\Delta T_B, \] and hence \[ \boxed{ T_A=T_0+\frac{Q}{m(c-g\alpha R_0)}, \qquad T_B=T_0+\frac{Q}{m(c+g\alpha R_0)}. } \] For ordinary conditions, \(g\alpha R_0\ll c\), giving the stated ordering.

Physically, A releases gravitational potential energy as it expands, whereas B uses some supplied energy to raise its centre of mass. No numerical temperature difference can be calculated from the supplied information.

The solution was checked against the official solution.