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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

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 \(B\).

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

BIO Shivering. Shivering is your body's way of generating heat to restore its internal temperature to the normal 37\(^\circ\)C, and it produces approximately 290 W of heat power per square meter of body area. A 68-kg, 1.78-m-tall woman has approximately 1.8 m\(^2\) of surface area. How long would this woman have to shiver to raise her body temperature by 1.0 C\(^\circ\), assuming that the body loses none of this heat? The body’s specific heat capacity is about 3500 J/kg \(\cdot\) K.

A 6.00-kg piece of solid copper metal at an initial temperature \(T\) is placed with 2.00 kg of ice that is initially at -20.0\(^\circ\)C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 1.20 kg of ice and 0.80 kg of liquid water. What was the initial temperature of the piece of copper?

BIO Conduction Through the Skin. The blood plays an important role in removing heat from the body by bringing this energy directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. Assume that the blood is brought to the bottom layer of skin at 37.0\(^\circ\)C and that the outer surface of the skin is at 30.0\(^\circ\)C. Skin varies in thickness from 0.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.75 mm. A 165-lb, 6-ft-tall person has a surface area of about 2.0 m\(^2\) and loses heat at a net rate of 75 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person’s skin?

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