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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Capitolo 17, Problema 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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Concetti chiave

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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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Percorso guidato
05:14
Overview of Heat Transfer

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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Percorso guidato
10:40
Latent Heat & Phase Changes
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