Problem 64 - 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 ice in 10.0 \(\mathrm{min} .\) Find the thermal conductivity \(k\) of the metal.

Step-by-Step Solution

Answer

The thermal conductivity of the metal is approximately 9.08 W/m·K.

Step 1: Identify the Heat Transfer Equation

The heat conducted through a rod can be calculated using the formula: \[ Q = \frac{kA(T_1 - T_2)t}{L} \] where \( Q \) is the heat transferred, \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( T_1 \) and \( T_2 \) are the temperatures at each end of the rod, \( t \) is the time of heat transfer, and \( L \) is the length of the rod.

Step 2: Calculate the Heat Required to Melt Ice

The heat required to melt ice is given by \( Q = mL_f \), where \( m \) is the mass of the ice melted and \( L_f \) is the latent heat of fusion for ice. With \( m = 8.50 \) g \((0.0085 \text{ kg})\) and \( L_f = 334,000 \text{ J/kg} \), \( Q = 0.0085 \times 334,000 = 2839 \text{ J} \).

Step 3: Solve for Thermal Conductivity \(k\)

We have \( Q = 2839 \text{ J} \), \( A = 1.25 \text{ cm}^2 = 1.25 \times 10^{-4} \text{ m}^2 \), \( T_1 = 100^\circ C \), \( T_2 = 0^\circ C \), \( t = 600 \text{ s} \), and \( L = 60.0 \text{ cm} = 0.60 \text{ m} \). Rearrange the heat transfer equation to \( k = \frac{QL}{A(T_1 - T_2)t} \) and substitute the known values: \[ k = \frac{2839 \times 0.60}{1.25 \times 10^{-4} \times 100 \times 600} \approx 9.08 \text{ W/m} \cdot \text{K} \]

Step 4: Short Answer

The thermal conductivity of the metal is approximately 9.08 W/m·K.

Key Concepts

Heat Transfer

Heat transfer is the movement of thermal energy from a hotter object to a cooler one. This can occur through various means like conduction, convection, or radiation.
In the case of a metal rod, heat is transferred through conduction. This process involves the transfer of energy between particles or within a solid.
The key factors that influence the rate of heat transfer include:

• Temperature Difference (\(T_1 \) and \( T_2 \) in the rod example): Greater temperature differences lead to faster heat transfer.
• Material Properties (like thermal conductivity \( k \)): Metals generally have high thermal conductivities, meaning they transfer heat efficiently.
• Cross-sectional Area (\( A \)): Larger areas allow more heat to pass through.
• Time (\( t \)): The longer the duration, the more heat is transferred.
• Distance or Length (\( L \)): Longer distances lessen the rate of heat exchange.

Understanding how these factors affect heat transfer helps us in designing better thermal systems for various applications, such as insulation and engineering.

Latent Heat of Fusion

The latent heat of fusion is the amount of energy needed to change a substance from a solid to a liquid at its melting point without changing its temperature.
For ice, the latent heat of fusion is about 334,000 J/kg. This means 334,000 joules of energy are required to melt one kilogram of ice.
In our exercise, 8.50 grams of ice require a certain amount of energy to melt:

• First, convert grams to kilograms (8.50 g = 0.0085 kg).
• Then, calculate the energy using the formula \(Q = mL_f\), where \(Q\) is the energy, \(m\) is the mass, and \(L_f\) is the latent heat of fusion.
  
• The calculation yields \(Q = 0.0085 \times 334,000 = 2839\, J\).

This energy is crucial in understanding how much heat is absorbed without a temperature increase, allowing you to design systems that need controlled temperature environments.

Thermal Conduction Equation

The thermal conduction equation governs how heat moves through materials. It is expressed as \(Q = \frac{kA(T_1 - T_2)t}{L}\).
Each element in the equation serves a purpose in illustrating heat flow:

• \(Q\): The total heat transferred through the material.
• \(k\): The thermal conductivity, a measure of a material's ability to conduct heat. Higher values indicate better conductivity.
• \(A\): The area through which heat is passing, impacting the overall rate of conduction.
• \(T_1\) and \(T_2\): The temperature difference between the two points. A larger difference increases heat transfer.
• \(t\): How long the heat is being transferred, with longer times yielding more transfer.
• \(L\): The length of the path, affecting the rate at which heat moves from the hotter to the cooler section.

This equation is essential for solving practical problems like determining how long it will take a material to heat or cool and calculating the required insulation thickness to maintain desired temperatures.