A physicist studies the rate of heat transfer, H (in watts, \text{W}), through cylindrical… — ACT

A physicist studies the rate of heat transfer, $H$ (in watts, $\text{W}$ ), through cylindrical metal rods. The researcher determines that $H$ is directly proportional to both the cross-sectional area of the rod and the temperature difference ( $\Delta T$ , in kelvins, $\text{K}$ ) between its two ends, and inversely proportional to the length of the rod ( $L$ , in meters, $\text{m}$ ). For Rod 1, the radius is $0.020\text{ m}$ , the length is $0.80\text{ m}$ , the temperature difference is $50.0\text{ K}$ , and the rate of heat transfer is $100.0\text{ W}$ . Rod 2 is made of the same metal and has a radius of $0.040\text{ m}$ , a length of $0.40\text{ m}$ , and a temperature difference of $30.0\text{ K}$ . What is the rate of heat transfer, in watts, for Rod 2?

Answer: 480 W

Answer

The rate of heat transfer for Rod 2 is 480 W.

The correct answer is 480 W because the rate of heat transfer is proportional to the square of the radius and the temperature difference, and inversely proportional to the length. The radius is doubled (scaling factor of

2^2 = 4

), the temperature difference is multiplied by 0.6, and the length is halved (scaling factor of

\frac{1}{0.5} = 2

). This yields a combined factor of

4 \times 0.6 \times 2 = 4.8

, and multiplying 100.0 W by 4.8 results in 480 W.

Step-by-Step Solution

Relate the cross-sectional area to the radius of the rod.

The area

A

is directly proportional to the square of the radius

r

A \propto r^2

The cross-section of a cylinder is a circle with area

A = \pi r^2

Formulate the complete proportionality expression for the rate of heat transfer.

H \propto \frac{r^2 \cdot \Delta T}{L}

Heat transfer is directly proportional to area (

r^2

) and temperature difference (

\Delta T

), and inversely proportional to length (

L

Set up a ratio to compare Rod 2's heat transfer rate to Rod 1's rate.

\frac{H_2}{H_1} = \left(\frac{r_2}{r_1}\right)^2 \cdot \left(\frac{\Delta T_2}{\Delta T_1}\right) \cdot \left(\frac{L_1}{L_2}\right)

Using a ratio cancels out the constant of proportionality.

Calculate the ratio multiplier by inserting the known values.

Multiplier =

2^2 \cdot 0.6 \cdot 2 = 4 \cdot 0.6 \cdot 2 = 4.8

The radius doubles (factor of 4), the temperature difference is multiplied by 0.6, and the length is halved (factor of 2).

Multiply the original heat transfer rate by the calculated factor.

H_2 = 100.0 \text{ W} \times 4.8 = 480 \text{ W}

To find the final heat transfer rate of Rod 2.

Key Concept

Combining direct and inverse proportionalities to calculate a new value using scaling factors.

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