Question:

A metal rod of length 1 meter has a thermal conductivity of 150 W/m·K. The rod is kept between two heat reservoirs, A and B, maintained at temperatures of 500°C and 100°C respectively. The rod is insulated from the surroundings.

1. Calculate the rate of heat transfer through the rod if the cross-sectional area of the rod is 0.01 m².

2. Determine the temperature difference across the rod.

3. Calculate the rate at which heat is transferred from reservoir A to reservoir B.

Answer:

1. To calculate the rate of heat transfer through the rod, we can use the formula for heat conduction:

   $\displaystyle q=\dfrac{kA\Delta T}{L}$

   where:

   • $\displaystyle q$ is the rate of heat transfer through the rod,
   • $\displaystyle k$ is the thermal conductivity of the rod,
   • $\displaystyle A$ is the cross-sectional area of the rod,
   • $\displaystyle \Delta T$ is the temperature difference across the rod, and
   • $\displaystyle L$ is the length of the rod.

   Plugging in the given values:

   $\displaystyle q=\dfrac{\left( 150\, \text{W/m} \cdot \text{K}\right) \times \left( 0.01\, \text{m}^{2}\right) \times \left( 500\, \text{K}-100\, \text{K}\right) }{1\, \text{m}}$

   $\displaystyle q=\dfrac{1500\, \text{W} \cdot \text{K}}{1\, \text{m}}$

   $\displaystyle q=1500\, \text{W}$

   Therefore, the rate of heat transfer through the rod is $\displaystyle 1500$ W.

2. The temperature difference across the rod can be calculated using the formula for temperature difference in heat conduction:

   $\displaystyle \Delta T=\dfrac{q\cdot L}{k\cdot A}$

   Plugging in the given values:

   $\displaystyle \Delta T=\dfrac{\left( 1500\, \text{W}\right) \times \left( 1\, \text{m}\right) }{\left( 150\, \text{W/m} \cdot \text{K}\right) \times \left( 0.01\, \text{m}^{2}\right) }$

   $\displaystyle \Delta T=\dfrac{1500\, \text{WK}}{1.5\, \text{W}}$

   $\displaystyle \Delta T=1000\, \text{K}$

   Therefore, the temperature difference across the rod is $\displaystyle 1000$ K.

3. The rate at which heat is transferred from reservoir A to reservoir B can be calculated using the formula:

   $\displaystyle q_{\text{AB}}=\dfrac{kA\left( T_{\text{A}}-T_{\text{B}}\right) }{L}$

   Plugging in the given values:

   $\displaystyle q_{\text{AB}}=\dfrac{\left( 150\, \text{W/m} \cdot \text{K}\right) \times \left( 0.01\, \text{m}^{2}\right) \times \left( 500\, \text{K}-100\, \text{K}\right) }{1\, \text{m}}$

   $\displaystyle q_{\text{AB}}=\dfrac{1500\, \text{W} \cdot \text{K}}{1\, \text{m}}$

   $\displaystyle q_{\text{AB}}=1500\, \text{W}$

   Therefore, the rate at which heat is transferred from reservoir A to reservoir B is $\displaystyle 1500$ W.