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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 17, Problem 56b

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 0.300 m and the length of the copper section is 0.8000.800 m. Each segment has cross-sectional area 0.005000.00500 m2. 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. What mass of ice is melted in 5.005.00 min by the heat conducted by the composite rod?

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First, understand that the problem involves heat conduction through two different materials, brass and copper, which are joined end to end. The heat will flow from the hot end (boiling water) to the cold end (ice-water mixture).
Use the formula for heat conduction: \( Q = \frac{k \cdot A \cdot \Delta T \cdot t}{L} \), where \( Q \) is the heat transferred, \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, \( t \) is the time, and \( L \) is the length of the rod.
Calculate the heat conducted through each segment separately. For the brass segment, use its thermal conductivity \( k_{brass} \), length \( L_{brass} = 0.300 \) m, and the temperature difference between boiling water (100°C) and the junction temperature. For the copper segment, use \( k_{copper} \), length \( L_{copper} = 0.800 \) m, and the temperature difference between the junction temperature and the ice-water mixture (0°C).
Since the rods are in series, the heat conducted through the brass segment equals the heat conducted through the copper segment. Set up an equation to solve for the junction temperature, ensuring continuity of heat flow.
Once the heat \( Q \) is calculated, use the latent heat of fusion for ice \( L_f = 334,000 \) J/kg to find the mass of ice melted: \( m = \frac{Q}{L_f} \). This will give you the mass of ice melted in 5 minutes.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Thermal Conductivity

Thermal conductivity is a material property that indicates how well a material can conduct heat. It is crucial for calculating the rate of heat transfer through the rods, as different materials like brass and copper have distinct thermal conductivities. This concept helps determine the amount of heat transferred from the boiling water to the ice-water mixture.
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Heat Transfer Equation

The heat transfer equation, Q = kA(T_hot - T_cold)t/L, is used to calculate the amount of heat conducted through a material. Here, Q is the heat transferred, k is the thermal conductivity, A is the cross-sectional area, T_hot and T_cold are the temperatures at each end, t is the time, and L is the length of the rod. This equation is essential for determining the heat conducted by the composite rod.
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Latent Heat of Fusion

Latent heat of fusion is the amount of heat required to change a unit mass of a substance from solid to liquid at constant temperature. For ice, this value is crucial to calculate the mass of ice melted by the heat conducted through the rods. It allows us to relate the heat transferred to the physical change in the ice, providing the final answer to the question.
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Related Practice
Textbook Question

Suppose that the rod in Fig. 17.2417.24a is made of copper, is 45.045.0 cm long, and has a cross-sectional area of 1.251.25 cm2 . Let TH=100.0TH = 100.0°C and TC=0.0TC = 0.0°C. What is the final steady-state temperature gradient along the rod?

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Textbook Question

A carpenter builds an exterior house wall with a layer of wood 3.03.0 cm thick on the outside and a layer of Styrofoam insulation 2.22.2 cm thick on the inside wall surface. The wood has k=0.080W/mKk=0.080\,W/m\(\cdot\) K , and the Styrofoam has k=0.027W/mKk=0.027\,W/m\(\cdot\) K. The interior surface temperature is 19.019.0°C, and the exterior surface temperature is 10.0-10.0°C. What is the temperature at the plane where the wood meets the Styrofoam?

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Textbook Question

A vessel whose walls are thermally insulated contains 2.402.40 kg of water and 0.4500.450 kg of ice, all at 0.00.0°C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.028.0°C? You can ignore the heat transferred to the container.

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Textbook Question

A carpenter builds an exterior house wall with a layer of wood 3.03.0 cm thick on the outside and a layer of Styrofoam insulation 2.22.2 cm thick on the inside wall surface. The wood has k=0.080W/mKk=0.080\,W/m\(\cdot\) K , and the Styrofoam has k=0.027W/mKk=0.027\,W/m\(\cdot\) K. The interior surface temperature is 19.019.0°C, and the exterior surface temperature is 10.0-10.0°C. What is the rate of heat flow per square meter through this wall?

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Textbook Question

A 4.004.00-kg silver ingot is taken from a furnace, where its temperature is 750.0750.0°C, and placed on a large block of ice at 0.00.0°C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

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Textbook Question

An insulated beaker with negligible mass contains 0.2500.250 kg of water at 75.075.0°C. How many kilograms of ice at 20.0-20.0°C must be dropped into the water to make the final temperature of the system 40.040.0°C?

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