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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

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

Step by step solution

01

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.
02

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 \( T_{boiling} = 100^\circ C \), \( T_{ice} = 0^\circ C \).
03

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, \( A = 0.005 \text{ m}^2 \)
04

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:\[ 109 \cdot (100 - T) \cdot 0.8 = 401 \cdot T \cdot 0.3 \]\[ 87.2 (100 - T) = 120.3 T \]\[ 8720 - 87.2T = 120.3T \]\[ 8720 = 207.5T \]\[ T \approx 42.03^\circ C \]
05

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:\[ q = 401 \cdot 0.005 \cdot \frac{42.03 - 0}{0.8} \cdot 300 \]\[ q \approx 3156.5 \text{ J} \]
06

Determine Mass of Ice Melted

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

One experimental method of measuring an insulating material's thermal conductivity is to construct a box of the material and measure the power input to an electric heater inside the box that maintains the interior at a measured temperature above the outside surface. Suppose that in such an apparatus a power input of 180 W is required to keep the interior surface of the box 65.0 C\(^\circ\) (about 120 F\(^\circ\)) above the temperature of the outer surface. The total area of the box is 2.18 m\(^2\), and the wall thickness is 3.90 cm. Find the thermal conductivity of the material in SI units.

You have 1.50 kg of water at 28.0\(^\circ\)C in an insulated container of negligible mass. You add 0.600 kg of ice that is initially at -22.0\(^\circ\)C. Assume that no heat exchanges with the surroundings. (a) After thermal equilibrium has been reached, has all of the ice melted? (b) If all of the ice has melted, what is the final temperature of the water in the container? If some ice remains, what is the final temperature of the water in the container, and how much ice remains?

CP While painting the top of an antenna 225 m in height, a worker accidentally lets a 1.00-L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

The emissivity of tungsten is 0.350. A tungsten sphere with radius 1.50 cm is suspended within a large evacuated enclosure whose walls are at 290.0 K. What power input is required to maintain the sphere at 3000.0 K if heat conduction along the supports is ignored?

A carpenter builds a solid wood door with dimensions 2.00 m \(\times\) 0.95 m \(\times\) 5.0 cm. Its thermal conductivity is k = 0.120 W/m \(\cdot\) K. The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.8-cm thickness of solid wood. The inside air temperature is 20.0\(^\circ\)C, and the outside air temperature is -8.0\(^\circ\)C. (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 m on a side is inserted in the door? The glass is 0.450 cm thick, and the glass has a thermal conductivity of 0.80 W/m \(\cdot\) K. The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.

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