Question

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

Short answer

(a) The joint temperature is approximately 42.03°C. (b) About 9.45 grams of ice melts.

Step-by-step solution

Understand Thermal Conductivity

The problem involves a composite rod with two materials: brass and copper. We need to understand how heat conducts through them. The thermal conductivity equation is: \[ q = K \cdot A \cdot \frac{dT}{dx} \]where \( q \) is the heat transfer per unit time, \( K \) is the thermal conductivity, \( A \) is the cross-sectional area, \( dT \) is the temperature difference, and \( dx \) is the length.

Define Equilibrium Condition

In thermal equilibrium, the heat transferred through both rods is equal, \( q_{brass} = q_{copper} \). Set the heat flows through the brass and copper equal to each other:\[ K_{brass} \cdot \frac{T_{boiling} - T}{L_{brass}} = K_{copper} \cdot \frac{T - T_{ice}}{L_{copper}} \] where \( T_{boiling} = 100^\circ C \), \( T_{ice} = 0^\circ C \).

Use Known Values

Use the given values: - Length of brass, \( L_{brass} = 0.300 \text{ m} \)- Length of copper, \( L_{copper} = 0.800 \text{ m} \)- Thermal conductivity: \( K_{brass} = 109 \text{ W/m} \, ^\circ\text{C} \) and \( K_{copper} = 401 \text{ W/m} \, ^\circ\text{C} \)- Area, \( A = 0.005 \text{ m}^2 \)

Solve for Equilibrium Temperature

Plug values into the equation:\[ 109 \cdot \frac{100 - T}{0.300} = 401 \cdot \frac{T - 0}{0.800} \]Solve for \( T \). Rearrange and simplify:\[ 109 \cdot (100 - T) \cdot 0.8 = 401 \cdot T \cdot 0.3 \]\[ 87.2 (100 - T) = 120.3 T \]\[ 8720 - 87.2T = 120.3T \]\[ 8720 = 207.5T \]\[ T \approx 42.03^\circ C \]

Calculate Heat Transferred

Use heat flow equation to find total heat transferred in 5 mins:\[ q = K \cdot A \cdot \frac{dT}{dx} \cdot t \]Use copper section for calculation:\[ q = 401 \cdot 0.005 \cdot \frac{42.03 - 0}{0.8} \cdot 300 \]\[ q \approx 3156.5 \text{ J} \]

Determine Mass of Ice Melted

Calculate mass of ice melted using heat of fusion (\( L_f = 334,000 \text{ J/kg} \)):\[ m_{ice} = \frac{q}{L_f} = \frac{3156.5}{334,000} \approx 0.00945 \text{ kg} \]

Key concepts

Heat Transfer

Heat transfer is the process by which thermal energy is exchanged between physical systems, depending on the temperature and the medium used. There are multiple modes of heat transfer: conduction, convection, and radiation. In this exercise, we focus on conduction, which is the transfer of heat through a material without the movement of the material itself.
For conduction, the heat transfer rate can be calculated using the formula:

• \[ q = K \cdot A \cdot \frac{dT}{dx} \]
• where \( q \) is the heat transfer per unit of time (measured in watts), \( K \) is the thermal conductivity of the material, \( A \) is the cross-sectional area through which the heat is conducted, \( dT \) is the temperature difference across the material, and \( dx \) is the thickness or length of the material.

Understanding these components helps us assess how effectively different materials transfer heat. For example, copper and brass have different thermal conductivities, affecting how heat is transferred through each part of the composite rod.

Temperature Equilibrium

Temperature equilibrium is a concept where two connected bodies or parts of a system reach a consistent temperature, stopping any net flow of heat energy between them. When dealing with composite materials like brass and copper rods, as in the exercise, you identify the equilibrium point by ensuring the heat transfer through each section is equal.
In formula terms, achieving temperature equilibrium involves setting the heat flow rates through both materials equal to each other:

• \[ K_{brass} \cdot \frac{T_{boiling} - T}{L_{brass}} = K_{copper} \cdot \frac{T - T_{ice}}{L_{copper}} \]
• This ensures the heat entering and leaving the joint point is evenly distributed.

The exercise demonstrates this with an equilibrium temperature calculated around \(42.03^\circ C\). This temperature is the point at which heat transfer balances between the hot and cold ends of the rod.

Phase Change - Melting

Phase changes occur when a substance transitions from one state of matter to another due to changes in temperature or pressure. In this exercise, as heat is conducted through the copper rod to melt ice, a phase change from solid to liquid occurs, which is known as melting. The amount of heat required for a phase change is determined by the heat of fusion, specific to the substance.
For ice, the heat of fusion is \( 334,000 \text{ J/kg} \), which represents the energy required to convert 1 kg of ice at 0°C to 1 kg of water at 0°C. The heat transfer calculated in the exercise helps determine how much ice melts over a certain period:

• \[ m_{ice} = \frac{q}{L_f} \]
• where \( q \) is the heat transferred, and \( L_f \) is the heat of fusion.
• For the example given, about \( 0.00945 \text{ kg} \) of ice melts in 5 minutes.

Such calculations are crucial in understanding thermal management in systems involving phase changes.

Metals - Brass and Copper

Metals like brass and copper are widely used in applications involving heat due to their substantial thermal conductivity. Thermal conductivity represents how quickly and efficiently heat passes through a material. In this exercise, brass and copper rods are chosen for their differing heat conducting capabilities, showcasing an important principle in thermal dynamics.
Each metal has distinct physical properties impacting heat transfer:

• Brass has a thermal conductivity of \( 109 \text{ W/m}\,^\circ C \), making it a moderate conductor of heat, useful in scenarios needing controlled heat transfer.
• Copper, with a higher thermal conductivity of \( 401 \text{ W/m}\,^\circ C \), rapidly transfers heat, making it ideal for quick heat dissipation applications.

Understanding the properties of different metals allows engineers to design systems efficiently, selecting materials best suited for the desired heat transfer outcomes.