Question:
The heat flows through a rod of length 50 cm and area of cross-section 5 cm\(^2\). Its ends are respectively at 25°C and 125°C. The coefficient of thermal conductivity of the material of the rod is 0.092 kcal/m°C. The temperature gradient of the rod is:
The heat flows through a rod of length 50 cm and area of cross-section 5 cm\(^2\). Its ends are respectively at 25°C and 125°C. The coefficient of thermal conductivity of the material of the rod is 0.092 kcal/m°C. The temperature gradient of the rod is:
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The temperature gradient is simply the change in temperature divided by the length of the rod, and it describes how temperature changes along the rod.
Updated On: Apr 22, 2026
- 2°C/cm
- 20°C/cm
- 20°C/m
- 2°C/m
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The Correct Option is A
Solution and Explanation
Step 1: Use the heat conduction formula.
The heat flow \( Q \) through a rod is given by Fourierâs Law: \[ Q = \frac{k A \Delta T}{L} \] where: - \( k \) is the coefficient of thermal conductivity, - \( A \) is the cross-sectional area, - \( \Delta T \) is the temperature difference, and - \( L \) is the length of the rod.
Step 2: Calculate the Temperature Gradient.
The temperature gradient is given by: \[ \text{Temperature Gradient} = \frac{\Delta T}{L} \] Substitute the given values: \[ \Delta T = 125^\circ C - 25^\circ C = 100^\circ C \] \[ L = 50 \, \text{cm} = 0.5 \, \text{m} \] Thus, the temperature gradient is: \[ \text{Temperature Gradient} = \frac{100}{50} = 2 \, \text{°C/cm} \]
The heat flow \( Q \) through a rod is given by Fourierâs Law: \[ Q = \frac{k A \Delta T}{L} \] where: - \( k \) is the coefficient of thermal conductivity, - \( A \) is the cross-sectional area, - \( \Delta T \) is the temperature difference, and - \( L \) is the length of the rod.
Step 2: Calculate the Temperature Gradient.
The temperature gradient is given by: \[ \text{Temperature Gradient} = \frac{\Delta T}{L} \] Substitute the given values: \[ \Delta T = 125^\circ C - 25^\circ C = 100^\circ C \] \[ L = 50 \, \text{cm} = 0.5 \, \text{m} \] Thus, the temperature gradient is: \[ \text{Temperature Gradient} = \frac{100}{50} = 2 \, \text{°C/cm} \]
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