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A copper rod is 0.871 m long with a rectangular cross section of dimensions 0.0533 m by 0.0775 m . The ends of the rod are held at 58.3 ∘ C and 97.7 ∘ C , but the rod is otherwise insulated from its environment. Find the heat current (the amount of heat that flows per unit time) through the rod. The thermal conductivity of copper is 385 W / (m ⋅ K) .

Answer :

Final answer:

The heat current through the copper rod is calculated using Fourier's Law of Heat Conduction. With a thermal conductivity of 385 W/m·K, cross-sectional area of 0.00413125 m², and a temperature difference of 39.4 K across a length of 0.871 m, the heat current is approximately 18.46 Watts.

Explanation:

For calculating heat current through a copper rod we can use Fourier's Law of Heat Conduction .It states that the time rate of heat flow Q through a material is proportional to the negative gradient in the temperature and to the area through which the heat flows. The formula is:

Q = -k * A * (dT/dx)

Where:

Q is the heat current (Watts, W)

k is the thermal conductivity of the material (W/m·K)

A is the cross-sectional area (m²)

dT is the temperature difference between the two ends (K)

dx is the length of the material (m)

The given parameters are:

Length of the copper rod (dx) = 0.871 m

Cross-sectional area (A) = 0.0533 m * 0.0775 m

Temperature difference (dT) = 97.7 ℃ - 58.3 ℃

Thermal conductivity of copper (k) = 385 W/m·K

First, we need to convert the temperature difference to Kelvin. Since the difference in Celsius is the same as in Kelvin:

dT = 97.7 - 58.3 = 39.4 K

Next, we calculate the cross-sectional area:

A = 0.0533 m * 0.0775 m = 0.00413125 m²

Then we can calculate the heat current:

Q = -385 W/m·K * 0.00413125 m² * 39.4 K / 0.871 m

Q = -18.46 W (approx., the negative sign indicates the direction of heat flow from hot to cold)

The heat current through the copper rod is therefore approximately 18.46 Watts.

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Rewritten by : Jeany

Answer:

Heat current = 71.926J/sec

Explanation:

Detailed explanation and calculation is shown in the image below