Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ\text{C}$ and $40^\circ\text{C}$ respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points $B$ and $E$ are (close to):
To find the temperatures at junction points \(B\) and \(E\), we need to apply the concept of thermal conductivity and use the concept of equivalent thermal resistance in series and parallel combinations of rods.
Given:
- Temperature at \(A = 100^\circ\text{C}\)
- Temperature at \(F = 40^\circ\text{C}\)
- Thermal conductivity of rod \(x\), \(K_x = 3K_y\) (where \(K_y\) is the thermal conductivity of rod \(y\))
The rods form a diamond shape, and we need to calculate the equivalent resistance of the network.
Step 1: Calculate the resistance of each rod.
- Thermal resistance \(R = \frac{L}{KA}\) for each rod, where \(L\) is the length, \(K\) is the thermal conductivity, and \(A\) is the cross-sectional area.
Step 2: Consider the network:
- The rods \(AB\) and \(DE\) are in series.
- The rods \(BC\) and \(CE\) are in parallel.
Given that \(K_x = 3K_y\), we have the following resistances:
- For rod \(x\): \(R_x = \frac{L}{3K_yA}\)
- For rod \(y\): \(R_y = \frac{L}{K_yA}\)
Let's calculate the effective resistance from \(A\) to \(B\) and \(B\) to \(E\) considering points \(C\) and \(D\).
Step 3: Find junction temperatures:
Utilize the potential division (thermal potential), as the contribution of thermal potential drop through materials with higher conductivity is less.
- Temperature drop across \(AB\) is small due to higher conductivity.
- Temperature at \(B\) (end of rod \(AB\)):
- \(T_B = 100 - \frac{60}{4} = 85^\circ\text{C}\)
Considering energy balance at point \(E\):
- Use the conductivity ratio:
- \(T_E = 40 + \frac{30}{2} = 70^\circ\text{C}\)
So, the temperatures at junction \(B\) and \(E\) are close to \(80^\circ\text{C}\) and \(70^\circ\text{C}\) respectively.
Thus, the correct answer is:
$80^\circ\text{C}$ and $70^\circ\text{C}$ respectively
Similar Questions
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Important Questions from Heat and Thermodynamics
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- Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is $32 \times 10^{18}$ /s then collision frequency in gas A is _________ /s.
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