Question

A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other (Fig. E17.62). The rod consists of a 1.00-m section of copper (one end in boiling water) joined end to end to a length \(L_2\) of steel (one end in the ice-water mixture). Both sections of the rod have crosssectional areas of 4.00 cm\(^2\). The temperature of the copper- steel junction is 65.0\(^\circ\)C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length \(L_2\) of the steel section?

Short answer

a) Heat flow is 5.39 W. b) Length of steel is 1.92 m.

Step-by-step solution

Identify Known Variables

We are given the following information:- Length of copper section: \( L_1 = 1.00 \, \text{m} \)- Boiling water temperature: \( T_1 = 100 \, ^\circ\text{C} \)- Junction temperature: \( T_j = 65.0 \, ^\circ\text{C} \)- Ice-water mixture temperature: \( T_3 = 0 \, ^\circ\text{C} \)- Cross-sectional area: \( A = 4.00 \, \text{cm}^2 = 4.00 \times 10^{-4} \, \text{m}^2 \)

Calculate Heat Transfer Through Copper

The heat flow through a material is given by \[Q = \frac{k \cdot A \cdot \Delta T}{L}\]For copper, the thermal conductivity \(k_\text{copper} = 385 \, \text{W/m} \cdot \text{K}\). The temperature difference through the copper is \( \Delta T_\text{copper} = T_1 - T_j \). Therefore, the heat flow becomes:\[Q_\text{copper} = \frac{385 \, \cdot 4.00 \times 10^{-4} \, \cdot (100 - 65)}{1.00}\]Calculating this gives the heat flow through the copper.

Calculate Heat Transfer Through Steel

The heat flow through the steel section must be the same as through the copper in the steady state. Using the formula\[Q_\text{steel} = \frac{k \cdot A \cdot \Delta T}{L_2}\]where the thermal conductivity \(k_\text{steel} = 50 \, \text{W/m} \cdot \text{K}\) and \( \Delta T_\text{steel} = T_j - T_3 = 65 - 0 \), we set\[Q_\text{copper} = Q_\text{steel}\]Solve for \( L_2 \) in this expression, using the \(Q_\text{copper}\) value calculated in Step 2.

Solve for Length of Steel Section, \(L_2\)

Equating the heat flows through the copper and steel sections:\[\frac{385 \, \cdot 4.00 \times 10^{-4} \, \cdot 35}{1.00} = \frac{50 \, \cdot 4.00 \times 10^{-4} \, \cdot 65}{L_2}\]Rearrange to solve for \(L_2\):\[L_2 = \frac{50 \, \cdot 65}{385 \, \cdot 35}\]Calculate \( L_2 \) to find the length of the steel section.

Key concepts

Heat Transfer

Heat transfer is a vital concept in understanding how thermal energy moves between objects. It occurs when there is a temperature difference between two regions. Heat flows naturally from a warmer object to a cooler one. There are three main modes of heat transfer: conduction, convection, and radiation. In this exercise, we focus on conduction.

• Conduction: This is the direct transfer of heat through a material without the movement of the material itself. It occurs in solids, where vibrating particles transfer energy to neighboring particles.

The formula used to calculate the rate of heat transfer through conduction is: \[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]where

• \( Q \) is the heat transferred per unit time (W),
• \( k \) is the thermal conductivity of the material (W/m·K),
• \( A \) is the cross-sectional area (m²),
• \( \Delta T \) is the temperature difference (K), and
• \( L \) is the length over which the heat is transferred (m).

Understanding this formula is crucial because it helps predict how heat moves through different materials, as demonstrated in our copper and steel rod problem.

Steady State

The term "steady state" refers to a condition where the variables (such as temperature, pressure, and volume) in a physical system remain constant over time. In the context of thermal processes, it means the rate of heat entering one section of a material equals the rate of heat leaving that section. Therefore, the temperature at each point in the material does not change over time. Steady state is reached when the amount of heat flowing into a system component equals the amount being transferred out. For the rod problem:

• Heat enters the copper section from boiling water.
• Heat exits the steel section into the ice-water mixture.

When at steady state, the heat flow rate in the copper section equals the heat flow rate in the steel section. This balance allows us to solve for unknown variables, like the length of the steel rod, since all other conditions remain constant.

Thermal Contact

Thermal contact refers to how two objects in close proximity can transfer heat between them. It's an essential concept in thermodynamics as it impacts how efficiently heat conducts from one material to another. When objects are in perfect thermal contact, their surfaces are in complete contact without any air gaps. This allows maximal heat transfer between them. In practical scenarios:

• Good thermal contact means minimal resistance to heat flow, enhancing efficient heat transfer.
• Poor thermal contact, often due to surface roughness, air gaps or impurities, reduces efficiency.

In the rod exercise, the copper is in perfect thermal contact with the boiling water, and the steel with the ice-water mixture. This ensures that heat transfer occurs swiftly across the junction without significant loss.

Cross-Sectional Area

The cross-sectional area of a material is an important factor in determining the rate of heat transfer through that material by conduction. It refers to the area of the material through which heat is conducted.The formula for heat conduction, \( Q = \frac{k \cdot A \cdot \Delta T}{L} \), illustrates that the heat transfer rate is directly proportional to the area:

• A larger cross-sectional area allows more heat to flow at a given time.
• A smaller area restricts the amount of heat transferred.

This relationship is crucial in designing systems for thermal management. In the context of our exercise, both the copper and steel sections of the rod have the same cross-sectional area of 4.00 cm². This consistency ensures a uniform transfer rate across the materials, simplifying calculations and allowing us to focus on other variables, like thermal conductivity and length, when examining heat flow in the steady state.