Figure shows how four thermally insulated heat conducting rods of equal lengths and equal cross-sectional areas are connected between two heat reservoirs maintained at temperatures 100 °C and 0 °C. A is a thermally insulated heat reservoir which always remains at constant temperature θ. The thermal conductivities k₁, k₂ and k₃ of the rods are 10, 30 and 50 W m⁻¹K⁻¹ respectively. At the steady state, the temperature θ of the reservoir A is

2018 A/L Physics Past Paper · Question 40 (English Medium)

Four conducting rods — the steady temperature of reservoir A

Figure shows how four thermally insulated heat conducting rods of equal lengths and equal cross-sectional areas are connected between two heat reservoirs maintained at temperatures $100\ ^\circ\text{C}$ and $0\ ^\circ\text{C}$ . $A$ is a thermally insulated heat reservoir which always remains at constant temperature $\theta$ . The thermal conductivities $k_1, k_2$ and $k_3$ of the rods are $10, 30$ and $50\ \text{W m}^{-1}\text{K}^{-1}$ respectively. At the steady state, the temperature $\theta$ of the reservoir $A$ is

(1) $90\ ^\circ\text{C}$
(2) $85\ ^\circ\text{C}$
(3) $80\ ^\circ\text{C}$
(4) $75\ ^\circ\text{C}$
(5) $65\ ^\circ\text{C}$

Answer and explanation

Answer: (1) $90\ ^\circ\text{C}$

Given information

All the rods have the same length $l$ and the same area $A_c$ .

Three rods go from the $100\ ^\circ\text{C}$ reservoir to $A$ : $k_1 = 10$ , $k_2 = 30$ , $k_3 = 50\ \text{W m}^{-1}\text{K}^{-1}$ .

One rod goes from $A$ to the $0\ ^\circ\text{C}$ reservoir: $k_1 = 10$ .

What the question asks

We want: the temperature $\theta$ of reservoir $A$ in the steady state.

Relevant theory

An object is in steady state when the temperatures of all points in it are steady.

Therefore, K is the rate of flow of heat per unit cross sectional area and per unit temperature gradient.

Required equations

$\frac{Q}{t} = \frac{kA_c\,\Delta\theta}{l}$

$\left(\frac{Q}{t}\right)_{in} = \left(\frac{Q}{t}\right)_{out}$

Applying it to this question

First write the rate of heat flow into $A$ .

The three rods have the same length and the same area. So their $k$ values add.

$10 + 30 + 50 = 90$

$\left(\frac{Q}{t}\right)_{in} = \frac{A_c}{l}(10 + 30 + 50)(100 - \theta)$

$\tfrac{A_c}{l}\,90(100 - \theta)$

$\left(\frac{Q}{t}\right)_{in} = \frac{A_c}{l}\,90\,(100 - \theta)$

Reasoning and calculation

$\tfrac{A_c}{l}\,10\theta$

$\left(\frac{Q}{t}\right)_{out} = \frac{A_c}{l}\,10\,(\theta - 0)$

$90(100 - \theta) = 10\theta$

$9000 - 90\theta = 10\theta$

$90$

$\theta = \frac{9000}{100}$

$\theta = 90\ ^\circ\text{C}$ . This is option (1).

Checks

Check the units.

The unit of $k$ is $\text{W m}^{-1}\text{K}^{-1}$ . The unit of $A_c/l$ is m. The unit of $\Delta\theta$ is K.

$\text{W m}^{-1}\text{K}^{-1} \times \text{m} \times \text{K} = \text{W}$

Both sides are in W. So $A_c/l$ cancels.

$100 * (k1 + k2 + k3) / (k1 + k2 + k3 + k1)$

The rods coming in conduct 9 times better than the rod going out.

So $A$ must be close to the hot reservoir. $\theta = 90\ ^\circ\text{C}$ makes sense.

Conclusion

$\theta = 90\ ^\circ\text{C}$

$90(100 - \theta) = 10\theta$

So $\theta = 90\ ^\circ\text{C}$ .

The answer is option (1).

Rods side by side with the same length and area act like one rod. Just add their $k$ values.

In the steady state, the rate of heat flow in equals the rate out.

Exam tip

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