Problem 56 - College Physics

Question

\(\bullet\) 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 min. Find the thermal conductivity \(k\) of the metal.

Step-by-Step Solution

Answer

The thermal conductivity \( k \) is approximately 0.056 W/m·K.

Step 1: Understand the Problem

We need to find the thermal conductivity \( k \) of a metal rod given that it melts 8.50 g of ice in 10 minutes. We know the temperatures at both ends of the rod, its length, cross-sectional area, and the amount of ice melted.

Step 2: Define the Key Formula

Use the formula for heat transfer through conduction: \( Q = \frac{kA(T_1 - T_2)t}{L} \), where \( Q \) is the heat transferred, \( A \) is the cross-sectional area, \( T_1 \) and \( T_2 \) are temperatures at both ends, \( t \) is the time, and \( L \) is the length.

Step 3: Calculate Heat Required to Melt Ice

Calculate the heat \( Q \) required to melt the ice using \( Q = mL_f \), where \( m = 8.50 \) g and \( L_f = 334 \text{ J/g} \) is the latent heat of fusion for ice. Convert mass to kg: \( m = 0.0085 \) kg, so \( Q = 0.0085 \times 334 = 2.839 \) J.

Step 4: Rearrange the Formula to Solve for k

Rewrite the formula to solve for \( k \): \( k = \frac{Q L}{A(T_1 - T_2)t} \).

Step 5: Plug in the Values

Insert the known values into the rearranged formula. Convert the area to m²: \( A = 1.25 \times 10^{-4} \) m² and length to meters: \( L = 0.6 \) m. Time in seconds is \( 600 \) s. The temperature difference is \( 100^{\circ}\text{C} - 0^{\circ}\text{C} = 100\). Substituting these values gives: \( k = \frac{2.839 \text{ J} \times 0.6 \text{ m}}{1.25 \times 10^{-4} \text{ m}^2 \times 100 \times 600 \text{ s}} \).

Step 6: Perform the Calculation

Calculate the thermal conductivity \( k \): \( k = \frac{2.839 \times 0.6}{1.25 \times 10^{-4} \times 100 \times 600} \approx 0.056 \text{ W/m} \cdot \text{K} \).

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept in thermal physics. It describes the movement of thermal energy from one body or substance to another. This transfer occurs due to a temperature difference between the two objects. Heat always flows from the hotter object to the cooler one, striving for thermal equilibrium.
Heat transfer can occur in three main ways:

• Conduction: The transfer of heat through a material without the movement of the material itself.
• Convection: Involves the movement of fluids like gases or liquids, carrying heat with them.
• Radiation: Transfer of heat through electromagnetic waves, such as sunlight warming the Earth.

In the given exercise, we're interested in conduction. Specifically, how heat travels from the hotter end of a rod to the cooler end, melting ice on the way.

Latent Heat of Fusion

The latent heat of fusion plays a crucial role in phase transitions. It refers to the amount of heat energy required to change a substance from a solid to a liquid at a constant temperature. This process doesn't increase the temperature of the substance, instead it changes the state from solid to liquid.
For ice (water), the latent heat of fusion is approximately 334 J/g. This means to melt 1 gram of ice, 334 Joules of energy is needed. In the exercise, when 8.50 g of ice melts, the heat necessary is calculated by multiplying the mass of the ice by the latent heat of fusion: \[ Q = mL_f = 0.0085 \text{ kg} \times 334 \text{ J/g} = 2.839 \text{ J} \]

Conduction Formula

The conduction formula is a mathematical expression that quantifies heat transfer through a material. It is given as:\[ Q = \frac{kA(T_1 - T_2)t}{L} \]Here's what each symbol represents:

• Q: Total amount of heat transferred, in Joules (J).
• k: Thermal conductivity of the material, indicating how well the material can conduct heat.
• A: Cross-sectional area through which heat is transferred, in square meters (m²).
• \(T_1\) and \(T_2\): Initial and final temperatures, respectively.
• t: Time over which the transfer occurs, in seconds (s).
• L: Length of the material through which heat is conducted, in meters (m).

In the problem, this formula helps in finding the thermal conductivity, \( k \), of the rod, illustrating how effectively it can conduct heat from one end to the other.

Temperature Difference

The temperature difference, often denoted as \( \Delta T \) or \(T_1 - T_2\), is the driving force behind heat transfer. It is the difference in temperature between two points or objects that causes heat to flow from the warmer to the cooler region. The greater the temperature difference, the greater the potential for heat transfer.
In the given exercise, the temperature difference is between the heated end of the rod maintained at \(100^\circ \text{C}\) and the cooler end in contact with ice at \(0^\circ \text{C}\). Thus, the temperature difference \( \Delta T \) is:\[ \Delta T = 100^\circ \text{C} - 0^\circ \text{C} = 100^\circ \text{C} \]\Understanding the temperature difference is crucial as it directly influences the rate of heat transfer through the rod and impacts the melting process of the ice.