Question

Two containers A and B are connected by a conducting solid cylindrical rod of length $\frac{242}{7}$ cm and radius $\sqrt{8.3}$ cm. Thermal conductivity of the rod is 693 watt/mole-K. The container A contains two mole of oxygen gas and the container B contains four mole of helium gas. At time t = 0 temperature difference of the containers is 50°C, after what time (in seconds) temperature difference between them will be 25°C. Transfer of heat takes place through the rod only. Neglect radiation loss. Take R = 8.3 J/mole-K and $\pi = \frac{22}{7}$.

Answer 1

Approximately 3 seconds.

Answer 2

Solution:

We use the fact that the heat‐flow through the rod is given by

$$\frac{dQ}{dt}=\frac{kA}{L}\,(T_A-T_B)$$

and that the energy change in a container is

$$dQ=m\,C\,dT.$$

Since the two containers are connected only by the rod, the loss from A equals the gain by B. It is most convenient to write the temperatures in “symmetric form”. Write

$$T_A=T_f+\frac{\Delta T}{2}\quad,\quad T_B=T_f-\frac{\Delta T}{2}$$

so that the difference $\Delta T=T_A-T_B$ decays exponentially. One may show that

$$\frac{d(\Delta T)}{dt}=-\frac{kA}{L}\Bigl(\frac{1}{m_AC_A}+\frac{1}{m_BC_B}\Bigr)\Delta T.$$

Thus,

$$\Delta T=\Delta T_0\,e^{-t/\tau}\quad,\quad \tau=\left[\frac{kA}{L}\Bigl(\frac{1}{m_AC_A}+\frac{1}{m_BC_B}\Bigr)\right]^{-1}.$$

Assumption on Heat Capacities:

For simple ideal gases at constant volume (a common assumption in such problems), we take:

• For O₂ (diatomic) the molar heat capacity $C_{V,O_2}=\frac{5}{2}R$.
• For He (monatomic) the molar heat capacity $C_{V,He}=\frac{3}{2}R$.

So for the two containers:

$$m_AC_A=2\Bigl(\frac{5}{2}R\Bigr)=5R\quad,\quad m_BC_B=4\Bigl(\frac{3}{2}R\Bigr)=6R.$$

Then,

$$\frac{1}{m_AC_A}+\frac{1}{m_BC_B}=\frac{1}{5R}+\frac{1}{6R}=\frac{6+5}{30R}=\frac{11}{30R}.$$

Thus,

$$\tau=\frac{30R}{11}\,\frac{L}{kA}\,.$$

Geometry of the Rod:

• Length: $L=\frac{242}{7}\text{ cm}=\frac{242}{7}\times\frac{1}{100}=\frac{242}{700}$ m.

• Radius: $r=\sqrt{8.3}$ cm $\implies$ in meters $=\frac{\sqrt{8.3}}{100}$.

• Area:

$$A=\pi r^2=\pi\Bigl(\frac{\sqrt{8.3}}{100}\Bigr)^2=\pi\,\frac{8.3}{10000} =\frac{22}{7}\,\frac{8.3}{10000}\,.$$

Substitution of Given Values:

We are given:

$$k=693\,\frac{\text{watt}}{\text{mole-K}},\quad R=8.3\,\frac{\text{J}}{\text{mole-K}}.$$

Thus,

$$\tau=\frac{30\times8.3}{11}\cdot\frac{L}{kA}.$$

Write $L=\frac{242}{700}$ m and

$$A=\frac{22\cdot8.3}{7\cdot10000}\,.$$

Combine:

$$\tau=\frac{30\cdot8.3}{11}\cdot\frac{\frac{242}{700}}{693\cdot\frac{22\cdot8.3}{7\cdot10000}}.$$

Notice that the factor $8.3$ cancels. Rearranging we get:

$$\tau=\frac{30\cdot242}{11\cdot700}\cdot\frac{7\cdot10000}{693\cdot22}.$$

Simplify step‐by‐step:

• $30\cdot242=7260$.
• $700$ in the denominator cancels with the factor $7$ as $7\times100=700$ so:

$$\tau=\frac{7260}{11\cdot700}\cdot\frac{7\cdot10000}{693\cdot22} =\frac{7260}{11\cdot700}\cdot\frac{10000}{693\cdot22/7}.$$

It is simpler to note that after canceling common factors one obtains a numerical value approximately:

$$\tau\approx 4.33\,\text{seconds.}$$

Since the temperature difference decays as:

$$\Delta T=\Delta T_0\,e^{-t/\tau},$$

with $\Delta T_0=50^\circ$C, when $\Delta T=25^\circ$C we have:

$$25=50\,e^{-t/\tau}\quad\Longrightarrow\quad e^{-t/\tau}=\frac{1}{2}\quad\Longrightarrow\quad t=\tau\,\ln2.$$

Taking $\ln2\approx0.693$:

$$t\approx 4.33\times0.693\approx 3\,\text{seconds}.$$

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Minimal Explanation:

1. Set up energy balance for two containers with $m_AC_A=5R$ and $m_BC_B=6R$.
2. Get differential equation for $\Delta T$: $$\frac{d\Delta T}{dt}=-\frac{kA}{L}\Bigl(\frac{1}{5R}+\frac{1}{6R}\Bigr)\Delta T.$$
3. Identify time constant: $$\tau=\frac{30R}{11}\,\frac{L}{kA}.$$
4. Substitute $R=8.3$, $L=\frac{242}{700}$ m, $A=\frac{22\cdot8.3}{7\cdot10000}$, $k=693$ to get $\tau\approx4.33$ s.
5. Since $25=50\,e^{-t/\tau}$ then $t=\tau\ln2\approx3$ s.