Question:

A metal rod of length 2m and thermal conductivity $K$ is initially at a uniform temperature of 100°C. One end of the rod is in contact with a furnace at 500°C while the other end is kept at 50°C. The furnace maintains a constant temperature throughout the experiment. The rod is made of a material with a specific heat capacity $C$ and a linear coefficient of thermal expansion $\alpha$.

a) Calculate the rate of heat transfer per unit time through the rod when it reaches a steady state.

b) Determine the length of the rod that will remain at a temperature of 200°C after reaching a steady state.

The given values are:

$K = 50 \, \text{W/mK}$ (thermal conductivity)

$C = 0.4 \, \text{J/g°C}$ (specific heat capacity)

$\alpha = 1.2 \times 10^{-5} \, \text{°C}^{-1}$ (linear coefficient of thermal expansion)

Answer:

a) To calculate the rate of heat transfer per unit time through the rod when it reaches a steady state, we need to use Fourier's law of heat conduction, which states that the rate of heat transfer (Q) through a material is proportional to the temperature difference across it and the thermal conductivity of the material.

The formula for the rate of heat transfer per unit time (Q/t) is given by:

$Q/t = -K * A * \frac{dT}{dx}$

Where:

• $-K$ is the thermal conductivity of the material (given as 50 W/mK)
• $A$ is the cross-sectional area perpendicular to the direction of heat transfer (which remains constant)
• $\frac{dT}{dx}$ is the temperature gradient across the rod

To find the rate of heat transfer per unit time, we need to calculate the temperature gradient ($\frac{dT}{dx}$) first.

Given that the rod has a length of 2m and a temperature difference of 500°C - 50°C = 450°C across it, we can calculate the temperature gradient as follows:

$\frac{dT}{dx} = \frac{T_{\text{hot end}} - T_{\text{cold end}}}{L_{\text{rod}}}$

$\frac{dT}{dx} = \frac{500 \, \text{°C} - 50 \, \text{°C}}{2 \, \text{m}}$

$\frac{dT}{dx} = 225 \, \text{°C/m}$

Now, plug in the values into the formula for the rate of heat transfer per unit time:

$Q/t = -K * A * \frac{dT}{dx}$

Since we are given the length of the rod (2m) but not the cross-sectional area, we can assume a circular cross-sectional area with a radius of $r$.

$A = \pi r^2$

Let's assume a radius of 0.01m for simplicity.

$A = \pi * (0.01 \, \text{m})^2 = 0.000314 \, \text{m}^2$

Plugging in the values:

$Q/t = -(50 \, \text{W/mK}) \cdot (0.000314 \, \text{m}^2) \cdot (225 \, \text{°C/m})$

$Q/t = -3.507 \, \text{W}$

So, the rate of heat transfer per unit time through the rod when it reaches a steady state is approximately -3.507 W (note the negative sign indicates heat is being transferred from the hot end to the cold end).

b) To determine the length of the rod that will remain at a temperature of 200°C after reaching a steady state, we can use the concept of thermal equilibrium.

At steady state, the rate of heat transfer into a section of the rod must be equal to the rate of heat transfer out of it. This means that the rate of heat transfer into the section between 200°C and 500°C must be equal to the rate of heat transfer out of the section between 200°C and 50°C.

The formula for the rate of heat transfer per unit time is the same as before:

$Q/t = -K * A * \frac{dT}{dx}$

We already know the rate of heat transfer per unit time through the rod (-3.507 W). Let's denote it as $Q/t = -3.507 \, \text{W}$.

Let's assume the length of the section that remains at a temperature of 200°C is $x$ meters.

At steady state, the temperature gradient within this section must be equal to the temperature gradient across the rod.

$\frac{dT}{dx} = \frac{dT_{\text{hot end}}}{dx} = \frac{T_{\text{hot end}} - T_{\text{cold end}}}{L_{\text{rod}}}$

Plugging in the values, we have:

$\frac{dT}{dx} = \frac{500 \, \text{°C} - 50 \, \text{°C}}{2 \, \text{m}} = 225 \, \text{°C/m}$

Now, we can set up the equation for the rate of heat transfer out of the section between 200°C and 50°C:

$Q/t = -K * A * \frac{dT}{dx} = -K * A * 225 \, \text{°C/m}$

Since the rate of heat transfer per unit time is the same for the entire rod, we can equate the two expressions for the rate of heat transfer:

$(-3.507 \, \text{W}) = -K * A * 225 \, \text{°C/m}$

We know the values of $K$ and $A$ from part a. Plugging in the values, we can solve for the length of the section that remains at a temperature of 200°C:

$(-3.507 \, \text{W}) = -(50 \, \text{W/mK}) * (0.000314 \, \text{m}^2) * 225 \, \text{°C/m} * x$

$x = \frac{-3.507 \, \text{W}}{-(50 \, \text{W/mK}) * (0.000314 \, \text{m}^2) * 225 \, \text{°C/m}}$

$x \approx 0.044 \, \text{m}$

Therefore, the length of the rod that will remain at a temperature of 200°C after reaching a steady state is approximately 0.044 meters.