Heat Transfer

To solve this problem, we will use the formula for the rate of heat flow through a cylindrical rod in steady state:

Q/t = (kAΔT)/L

Where:

• Q/t is the rate of heat flow.
• k is the thermal conductivity of the material.
• A is the cross-sectional area of the rod.
• ΔT is the temperature difference across the rod.
• L is the length of the rod.

Since the rods A and B have the same material and mass, their densities and volumes are the same, implying equal cross-sectional areas for equal length rods since they are cylindrical. Thus, the cross-sectional area A can be considered equal for both rods.

For rod A:

Q_A/t = (kAΔT_A)/L_A

For rod B:

Q_B/t = (kAΔT_B)/L_B

Given:

• ΔT_A = 40°C
• ΔT_B = 60°C
• Q_A/t : Q_B/t = 3 : 8

To find L_A/L_B}, we set up the ratio using the heat flow formula:

(Q_A/t) / (Q_B/t) = (ΔT_A * L_B) / (ΔT_B * L_A)

Substituting the given values:

3/8 = (40 * L_B) / (60 * L_A)

Simplifying the fractions, we get:

3/8 = (2/3) * (L_B/L_A)

Cross-multiplying to solve for the length ratio:

3 * 3 = 8 * 2 * (L_B/L_A)

9 = 16 * (L_B/L_A)

L_A/L_B = 16/9

Reversing the order for ratios:

L_B/L_A = 9/16

The ratio of the lengths of rods A to B is therefore:

4 : 3

(a)

Where:
- (moles of gas)
- (ideal gas constant)
- (temperature)
-

Thus, work done is:

Answer:

(b) The change in internal energy:
For an ideal gas undergoing an isothermal process, the change in internal energy ( ) is zero because internal energy of an ideal gas depends only on temperature, and the temperature does not change in an isothermal process.

Answer:

(c) The heat exchanged with the surroundings:
According to the first law of thermodynamics:

Since , we have:

Answer: