Question

During your mechanical engineering internship, you are given two uniform metal bars \(A\) and \(B\), which are made from different metals, to determine their thermal conductivities. Measuring the bars, you determine that both have length 40.0 cm and uniform cross-sectional area 2.50 cm\(^2\). You place one end of bar \(A\) in thermal contact with a very large vat of boiling water at 100.0\(^\circ\)C and the other end in thermal contact with an ice-water mixture at 0.0\(^\circ\)C. To prevent heat loss along the bar's sides, you wrap insulation around the bar. You weigh the amount of ice initially and find it to be 300 g. After 45.0 min has elapsed, you weigh the ice again and find that 191 g of ice remains. The ice-water mixture is in an insulated container, so the only heat entering or leaving it is the heat conducted by the metal bar. You are confident that your data will allow you to calculate the thermal conductivity \(k_A\) of bar \(A\). But this measurement was tedious-you don't want to repeat it for bar \(B\). Instead, you glue the bars together end to end, with adhesive that has very large thermal conductivity, to make a composite bar 80.0 m long. You place the free end of A in thermal contact with the boiling water and the free end of \(B\) in thermal contact with the ice-water mixture. As in the first measurement, the composite bar is thermally insulated. You go to lunch; when you return, you notice that ice remains in the ice-water mixture. Measuring the temperature at the junction of the two bars, you find that it is 62.4\(^\circ\)C. After 10 minutes you repeat that measurement and get the same temperature, with ice remaining in the ice-water mixture. From your data, calculate the thermal conductivities of bar \(A\) and of bar \(B\).

Short answer

Thermal conductivity of bar \(A\) is 21.44 W/m·K and for bar \(B\) is 43.89 W/m·K.

Step-by-step solution

Understanding the Scenario

We are trying to determine the thermal conductivities, \(k_A\) and \(k_B\), of two metal bars using the heat flow caused by a temperature gradient. Bar \(A\)'s thermal conductivity will be determined using changes in the ice mass melted, and the composite bar made by gluing \(A\) and \(B\) together will help us find \(k_B\) by measuring the steady state junction temperature.

Calculate Heat Transferred in Bar A

Initial mass of ice is 300 g, and the remaining is 191 g, so 109 g of ice is melted. The latent heat of fusion for ice is \(L = 333,000 \, \text{J/kg}\).\[Q = mL = 0.109 \, \text{kg} \times 333,000 \, \text{J/kg} = 36,297 \, \text{J}\]We assume this heat transfer happened over 45 minutes or 2700 seconds, so the rate of heat transfer \(\dot{Q}\) is \[\dot{Q} = \frac{36,297 \, \text{J}}{2700 \, \text{s}} = 13.4 \, \text{W}\].

Determine Thermal Conductivity for Bar A

Using the heat conduction formula: \[\dot{Q} = k_A \cdot A \cdot \frac{\Delta T}{L}\] where, \(\Delta T = 100.0 - 0.0 = 100 \, ^\circ \text{C}\), \(L = 0.4 \, \text{m}\), and \(A = 2.5 \times 10^{-4} \, \text{m}^2\). Solve for \(k_A\):\[k_A = \frac{13.4 \, \text{W} \times 0.4 \, \text{m}}{2.5 \times 10^{-4} \, \text{m}^2 \times 100 \, ^\circ \text{C}} = 21.44 \, \text{W/m}\cdot\text{K}\].

Analyze the Junction Temperature for Composite Bar

For the composite bar with steady junction temperature \(T_j = 62.4 \, ^\circ \text{C}\), heat flow in both sections must be equal: \(\dot{Q}_A = \dot{Q}_B\). The formulas for heat conduction in bar \(A\) and \(B\) are: \[\dot{Q}_A = k_A \cdot A \cdot \frac{T_j - T_1}{L_A}\] and \[\dot{Q}_B = k_B \cdot A \cdot \frac{T_2 - T_j}{L_B}\] with respective lengths and temperature differences.

Calculate Thermal Conductivity of Bar B

Setting \(\dot{Q}_A\) equal to \(\dot{Q}_B\):\[k_A \cdot \frac{T_j - 100}{0.4} = k_B \cdot \frac{0 - T_j}{0.4}\]Solve for \(k_B\) using previously found \(k_A\):\[21.44 \cdot \frac{62.4 - 100}{0.4} = k_B \cdot \frac{0 - 62.4}{0.4}\]\[k_B = 43.89 \, \text{W/m}\cdot\text{K}\].

Key concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one substance to another. It occurs due to a difference in temperature between objects. In this exercise, the heat transfer is taking place between the metal bars and the surrounding boiling water and ice-water mixture.

There are three common modes of heat transfer:

• Conduction: This is the transfer of heat through a material without the movement of the material itself. It involves the transfer of energy through particle collisions and is the primary mode considered in this exercise, as evidenced by the use of metal bars.
• Convection: This involves the movement of heat through a fluid (liquid or gas) caused by the fluid's motion. In this case, it is not directly considered, as the problem setup prevents extensive fluid movements.
• Radiation: This is the emission of energy as electromagnetic waves. Although surrounding factors may involve radiation, the problem focuses on conduction.

Heat transfer in the metal bar is exclusively due to conduction. This means the thermal energy flows from the hot end of the bar, exposed to boiling water, to the cold end, at the ice-water mixture, as determined by their temperature difference.

Latent Heat of Fusion

Latent heat of fusion refers to the amount of heat needed to change a substance from solid to liquid without changing its temperature. For ice turning into water, this energy is represented as latent heat of fusion.

This concept is crucial in our exercise because melting ice helps in quantifying the heat transferred through the metal bar. As given, the latent heat of fusion for ice is 333,000 J/kg. Thus, by knowing the mass of ice that melted, we can calculate the total heat absorbed by the ice.

Let's recap the formula to find the energy used:

• Heat absorbed by the melting ice ( Q ) is given by Q = m imes L , where