Question

One end of an insulated metal rod is maintained at 100.0\(^\circ\)C, and the other end is maintained at 0.00\(^\circ\)C by an ice-water mixture. The rod is 60.0 cm long and has a cross-sectional area of 1.25 cm\(^2\). The heat conducted by the rod melts 8.50 g of ice in 10.0 min. Find the thermal conductivity \(k\) of the metal.

Short answer

Thermal conductivity \(k\) is 22.71 W/m·K.

Step-by-step solution

Understand the Problem

We need to find the thermal conductivity \(k\) of the metal rod given the amount of ice melted by heat conduction through the rod over a specific time period. We have the rod's temperature difference, dimensions, and the mass of ice melted.

Use the Heat Conducted Formula

The formula for heat conducted through a rod is \(Q = k \cdot A \cdot \Delta T \cdot \frac{t}{L}\), where \(Q\) is the heat transferred, \(A\) is the cross-sectional area, \(\Delta T\) is the temperature difference, \(t\) is the time, and \(L\) is the length of the rod.

Calculate the Heat Required to Melt the Ice

The heat required to melt ice is given by \(Q = m \cdot L_f\), where \(m\) is the mass of the ice and \(L_f\) is the latent heat of fusion of ice (334 J/g). For 8.50 g of ice, \(Q = 8.50 \, \text{g} \times 334 \, \text{J/g} = 2839 \, \text{J}\).

Relate to the Conducted Heat Formula

We substitute the known values into the heat conducted formula \(Q = k \cdot A \cdot \Delta T \cdot \frac{t}{L}\) and solve for \(k\). Given \(Q = 2839 \, \text{J}\), \(A = 1.25 \, \text{cm}^2 = 1.25 \times 10^{-4} \, \text{m}^2\), \(\Delta T = 100.0 \, ^\circ\text{C}\), \(t = 10.0 \, \text{min} = 600 \, \text{s}\), and \(L = 60.0 \, \text{cm} = 0.60 \, \text{m}\).

Solve for Thermal Conductivity \(k\)

Rearrange the formula to find \(k\):\[k = \frac{Q \cdot L}{A \cdot \Delta T \cdot t} = \frac{2839 \, \text{J} \times 0.60 \, \text{m}}{1.25 \times 10^{-4} \, \text{m}^2 \times 100.0 \, ^\circ\text{C} \times 600 \, \text{s}}.\]Calculate to find \(k\).

Perform the Calculation

Substitute the values into the equation:\[k = \frac{2839 \, \text{J} \times 0.60 \, \text{m}}{1.25 \times 10^{-4} \, \text{m}^2 \times 100.0 \, ^\circ\text{C} \times 600 \, \text{s}} = 22.71 \, \text{W/m} \cdot \text{K}.\]

Key concepts

Heat Transfer

Heat transfer is the process through which thermal energy is exchanged between two systems or bodies at different temperatures. In the context of the metal rod exercise, heat is transferred from the warmer end at 100.0\(^\circ\)C to the colder end in contact with the ice-water mixture, which is maintained at 0.0\(^\circ\)C.

This transfer continues until thermal equilibrium is reached, or as long as the temperature difference is maintained. The metal rod acts as a conduit for heat flow, showcasing the principle that heat moves from higher to lower temperatures. In practical applications, understanding heat transfer helps us design efficient systems for maintaining desired temperatures in engineering and everyday scenarios.

• Temperature differential is essential for heat transfer to occur.
• Heat always travels from hot to cold areas.
• The rate of transfer depends on material properties like thermal conductivity.

Latent Heat of Fusion

The latent heat of fusion is the amount of heat absorbed or released by a substance as it changes state from solid to liquid or vice versa at constant temperature. In our exercise with the metal rod, the latent heat of fusion relates to the energy required to melt the ice.

For ice, the latent heat of fusion is 334 J/g, meaning it takes 334 joules to melt one gram of ice without changing its temperature. The formula \(Q = m \cdot L_f\) calculates this energy needed, where \(m\) is the mass of ice and \(L_f\) is the latent heat of fusion.

• This concept is critical in calculating energy transitions in phase changes.
• The heat required does not cause a temperature change but changes ice to water.
• Understanding latent heat allows for better energy management in systems.

Temperature Gradient

The temperature gradient is the rate of temperature change with respect to distance through a material. It is a central concept in calculating heat transfer efficiency because a higher gradient typically indicates a faster heat transfer.

In the context of the insulated metal rod, the temperature gradient exists between the high temperature at one end (100.0\(^\circ\)C) and the low temperature at the other end (0.0\(^\circ\)C). This creates a consistent drive for heat to flow along the rod.

• The temperature gradient drives thermal energy movement.
• Larger gradients can lead to faster heat conduction.
• Gradients are essential in solving problems involving heat exchange.

Thermal Conductivity Calculation

Thermal conductivity is a measure of a material's ability to conduct heat. It is denoted by \(k\) and is crucial for determining how quickly heat can pass through a substance.

In our example, thermal conductivity calculation involves using the rearranged formula \(k = \frac{Q \cdot L}{A \cdot \Delta T \cdot t}\). Here, \(Q\) represents the heat transferred, \(L\) is the length of the rod, \(A\) is its cross-sectional area, \(\Delta T\) is the temperature difference, and \(t\) is the time duration.

Substituting the given values, the calculation results in \(k\) for the metal rod. This calculation is vital for designing and understanding systems where controlled heat transfer is necessary, such as in insulations and heat exchangers.

• High \(k\) values indicate better conductor of heat.
• The material's physical properties directly affect \(k\).
• Calculated \(k\) helps optimize thermal system designs.