Problem 66 - University Physics with Modern Physics

Question

One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C}\) and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) 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 ic. 0 min. Find the thermal conductivity \(k\) of the metal.

Step-by-Step Solution

Answer

The thermal conductivity \( k \) is approximately \( 227.1 \text{ W/m°C} \).

Step 1: Understand the problem

The problem involves calculating the thermal conductivity \( k \) of a metal rod given specific conditions. The temperatures at each end of the rod are different, there is a temperature difference \( \Delta T = 100.0^{\circ} \mathrm{C} - 0.00^{\circ} \mathrm{C} = 100.0^{\circ} \mathrm{C} \). The length \( L \) of the rod is 60.0 cm (or 0.60 meters), and the cross-sectional area \( A \) is 1.25 cm\(^2\). The heat conducted melts 8.50 g of ice, which helps us find the heat transfer \( Q \) over time \( t = 10 \) minutes.

Step 2: Convert units

Convert the given measurements to standard SI units for calculations. - Length \( L = 60.0 \text{ cm} = 0.60 \text{ m} \)- Cross-sectional area \( A = 1.25 \text{ cm}^2 = 1.25 \times 10^{-4} \text{ m}^2 \)- Mass of ice \( m = 8.50 \text{ g} = 0.00850 \text{ kg} \)

Step 3: Calculate heat energy

The amount of heat \( Q \) required to melt the ice is given by \( Q = mL_f \), where \( L_f \) is the latent heat of fusion for ice, \( L_f = 334,000 \text{ J/kg} \). \[ Q = 0.00850 \times 334,000 = 2839 \text{ J} \]

Step 4: Calculate heat transfer rate

Since the heat transfer is given over 10 minutes, convert the time to seconds: \( t = 10 \times 60 = 600 \) seconds.The rate of heat transfer \( \frac{Q}{t} \) can be calculated as: \[ \frac{Q}{t} = \frac{2839}{600} = 4.732 \text{ W} \]

Step 5: Apply Fourier's Law

Fourier's Law for heat conduction states: \[ \frac{Q}{t} = kA \frac{\Delta T}{L} \]We need to solve this equation for \( k \), the thermal conductivity:Rearranging, \[ k = \frac{Q}{t} \frac{L}{A \Delta T} \]

Step 6: Calculate thermal conductivity

Substitute the known values into the rearranged Fourier's Law:\[ k = \frac{4.732 \text{ W} \times 0.60 \text{ m}}{1.25 \times 10^{-4} \text{ m}^2 \times 100.0 ^{\circ} \text{C}} \]\[ k = \frac{2.8392}{1.25 \times 10^{-2}} = 227.136 \text{ W/m°C} \]

Step 7: Conclusion

The thermal conductivity \( k \) of the rod is approximately \( 227.1 \text{ W/m°C} \).

Key Concepts

Fourier's Law

As you delve into the world of heat transfer, one of the most important laws you'll encounter is Fourier's Law of Heat Conduction. This principle provides a cornerstone for understanding how heat moves through materials. Fourier's Law can be expressed as \( \frac{Q}{t} = kA \frac{\Delta T}{L} \), where:

• \( Q \) is the heat energy transferred per unit time (watts, W)
• \( t \) is the time duration of heat transfer (seconds, s)
• \( k \) is the thermal conductivity of the material (W/m°C)
• \( A \) is the cross-sectional area through which heat conducts (m²)
• \( \Delta T \) is the temperature difference across the material (°C)
• \( L \) is the length of the material (m)

In simple terms, Fourier's Law states that the heat transfer rate through a material is proportional to the negative gradient of temperatures and the area through which heat flows. The thermal conductivity \( k \) is a property of the material that reflects its ability to conduct heat. The larger the value of \( k \), the better a material is at conducting heat. This law helps engineers and scientists design thermal systems and understand heat flow in various applications.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics and plays a crucial role in diverse areas from household heating systems to industrial processing. It's the movement of thermal energy due to a temperature difference within a physical system or between systems. There are three primary modes through which heat can be transferred:

• Conduction: The transfer of heat within a body or between two bodies in direct contact. Fourier's Law, which we discussed, describes this mode of heat transfer.
• Convection: The transfer of heat by the movement of fluids (liquids or gases). This involves the bulk movement of molecules within fluids.
• Radiation: The transfer of energy through electromagnetic waves, such as heat from the sun, without the need for a medium.

When you conduct experiments or solve problems involving heat transfer, identifying the mode of transfer is crucial as it influences the equations and constants you'll use. In the context of a metal rod, conduction is the primary mechanism at play, where heat travels from the warmer end to the cooler end.

Latent Heat of Fusion

Latent heat of fusion is an essential concept when dealing with phase changes. It represents the amount of heat energy needed to change a unit mass of a substance from a solid to a liquid without changing its temperature. For instance, when you melt ice, you need to provide a specific amount of energy to convert the solid ice to liquid water, even though the temperature remains at 0°C until the phase change is complete. The formula used to calculate the heat required for a phase change is:\[ Q = mL_f \]where:

• \( Q \) is the heat energy (joules, J)
• \( m \) is the mass of the substance (kilograms, kg)
• \( L_f \) is the latent heat of fusion (J/kg)

For water, the latent heat of fusion is approximately 334,000 J/kg. This value is pivotal in problems where ice or other substances undergo melting, as it links the mass of the substance to the energy required for its phase transition. Understanding this concept helps to accurately measure heat transfer in systems involving melting or freezing processes.