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The two ends of a rod of length L and a ...

The two ends of a rod of length `L` and a uniform cross-sectional area `A` are kept at two temperature `T_(1)` and `T_(2)` `(T_(1) gt T_(2))`. The rate of heat transfer. `(dQ)/(dt)`, through the rod in a steady state is given by

A

`(dQ)/(dt) = (KL (T_(1) - T_(2)))/(A)`

B

`(dQ)/(dt) = (L(T_(1) - T_(2)))/(LA)`

C

`(dQ)/(dt) = KLA (T_(1) - T_(2))`

D

`(dQ)/(dt) = (KA (T_(1) - T_(2)))/(L)`

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To solve the problem of determining the rate of heat transfer through a rod with given temperatures at its ends, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Setup**: - We have a rod of length \( L \) and uniform cross-sectional area \( A \). - One end of the rod is at temperature \( T_1 \) and the other end is at temperature \( T_2 \) with \( T_1 > T_2 \). 2. **Identify the Heat Transfer Mechanism**: - Heat transfer through the rod occurs due to conduction, which is governed by Fourier's law of heat conduction. 3. **Apply Fourier's Law**: - According to Fourier's law, the rate of heat transfer \( \frac{dQ}{dt} \) through a material is proportional to the temperature difference across it and the area through which heat is being transferred, and inversely proportional to the length of the rod. - The formula is given by: \[ \frac{dQ}{dt} = k \cdot A \cdot \frac{(T_1 - T_2)}{L} \] - Here, \( k \) is the thermal conductivity of the material of the rod. 4. **Substitute the Known Values**: - Since \( T_1 > T_2 \), the temperature difference \( (T_1 - T_2) \) is positive, which indicates that heat flows from the higher temperature end to the lower temperature end. 5. **Final Expression**: - Thus, the rate of heat transfer through the rod can be expressed as: \[ \frac{dQ}{dt} = k \cdot A \cdot \frac{(T_1 - T_2)}{L} \] ### Final Answer: \[ \frac{dQ}{dt} = k \cdot A \cdot \frac{(T_1 - T_2)}{L} \]
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