Problem 56 - College Physics

Question

One end of an insulated metal rod is maintained at \(100^{\circ} \mathrm{C}\). while the other end is maintained at \(0^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is \(60.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.25 \mathrm{~cm}^{2}\). The heat conducted by the rod melts \(8.50 \mathrm{~g}\) of ice in \(10.0 \mathrm{~min}\). Find the thermal conductivity \(k\) of the metal.

Step-by-Step Solution

Answer

The thermal conductivity \( k \) of the metal is approximately 22.71 W/m·K.

Step 1: Identify Known Values

The length of the rod \( L \) is given as \( 60.0 \text{ cm} = 0.6 \text{ m} \).The cross-sectional area \( A \) is \( 1.25 \text{ cm}^2 = 1.25 \times 10^{-4} \text{ m}^2 \).The temperature difference \( \Delta T \) is \( 100^{\circ}C - 0^{\circ}C = 100^{\circ}C \).The mass of ice melted \( m \) is \( 8.50 \text{ g} = 0.0085 \text{ kg} \).The time \( t \) is \( 10.0 \text{ minutes} = 600 \text{ seconds} \).The latent heat of fusion for ice \( L_f \) is \( 334,000 \text{ J/kg} \).

Step 2: Calculate the Heat Needed to Melt the Ice

The heat \( Q \) needed to melt the ice is given by the formula \( Q = m \cdot L_f \).Substituting the known values, we have:\[ Q = 0.0085 \times 334,000 = 2,839 \text{ J} \].

Step 3: Apply the Heat Transfer Equation

The formula for heat transfer through a material is \( Q = \frac{k \cdot A \cdot \Delta T \cdot t}{L} \), where \( k \) is the thermal conductivity.Rearrange this equation to solve for \( k \):\[ k = \frac{Q \cdot L}{A \cdot \Delta T \cdot t} \].

Step 4: Substitute Values into the Formula

Substitute the values into the equation from Step 3:\[ k = \frac{2,839 \text{ J} \times 0.6 \text{ m}}{1.25 \times 10^{-4} \text{ m}^2 \times 100 \text{ K} \times 600 \text{ s}} \].

Step 5: Perform the Calculation

Calculate \( k \) using the values from Step 4:\[ k = \frac{1,703.4}{75} = 22.71 \text{ W/m} \cdot \text{K} \].

Key Concepts

Heat Conduction

Heat conduction is the transfer of thermal energy through a material without any movement of the material itself. This process occurs due to the temperature difference between two ends of the material. In our exercise, the metal rod has one end at \(100^{\circ} \text{C}\) and the other at \(0^{\circ} \text{C}\), facilitating heat flow from the warmer to the cooler end. The rate of heat transfer depends on several factors:

• The thermal conductivity \(k\), which is a property specific to the material—higher conductivity means better heat transfer.
• The cross-sectional area \(A\) of the rod, as a larger area allows more heat to flow.
• The temperature difference \(\Delta T\) across the rod.
• The length \(L\) of the rod, where longer rods reduce the rate of heat flow.

The formula to calculate the heat transferred through conduction over time \(t\) is \(Q = \frac{k \cdot A \cdot \Delta T \cdot t}{L}\). In this exercise, students use this formula to comprehend how physical properties and dimensions affect heat transfer.

Latent Heat of Fusion

The latent heat of fusion is the amount of heat required to change a substance from solid to liquid at its melting point, without increasing its temperature. For ice, this value is \(334,000 \text{ J/kg}\). This concept is crucial to understanding the problem, as the heat conducted by the metal rod melts ice at \(0^{\circ} \text{C}\).When calculating the heat \(Q\) required to melt the ice, you use the formula \(Q = m \cdot L_f\), where \(m\) is the mass of the ice and \(L_f\) is the latent heat of fusion. In the exercise:

• The mass of the ice is \(8.50 \text{ g} = 0.0085 \text{ kg}\).
• Thus, \(Q = 0.0085 \times 334,000 = 2,839 \text{ J}\).

This calculation shows how much energy is needed to convert the ice into water entirely, which corresponds to the heat transferred through the metal rod.

Temperature Difference

Temperature difference is the primary driving force for heat conduction. It is the difference in temperature across the two ends of a material that causes heat to move from the warmer region to the cooler one. In the given problem, the ends of the rod have temperatures of \(100^{\circ} \text{C}\) and \(0^{\circ} \text{C}\), resulting in a temperature difference \(\Delta T = 100 \text{ C or } 100 \text{ K}\).This temperature gradient is crucial because it determines the rate at which heat flows through the rod. A greater temperature difference results in a higher rate of heat transfer. Keep in mind that this difference remains constant in the exercise, ensuring consistent conditions for calculating the thermal conductivity. Understanding the influence of temperature difference helps in grasping why heat moves and how it can be calculated using the heat conduction equation.

Calorimetry

Calorimetry is the science of measuring the amount of heat involved in chemical reactions or physical changes, such as melting ice. In the context of this problem, calorimetry principles help understand how much thermal energy from the rod is used to melt a known mass of ice. We can say calorimetry focuses on:

• Quantifying the heat required to achieve temperature changes or phase transitions.
• Using known heats of fusion or vaporization to track how energy is absorbed or released during these processes.

In the problem, calorimetry principles are applied by using the latent heat of fusion to ascertain how much heat is necessary to change the ice into water. Calculations based on these calorimetry concepts lead us to determine the thermal conductivity of the rod, ensuring students can make the connection between energy transfer and material properties.