Ch 17: Temperature and Heat

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 17: Temperature and Heat

Problem 54b

Chapter 17, Problem 54b

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 $0.00500$ 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 $5.00$ min by the heat conducted by the composite rod?

Verified step by step guidance

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.

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.

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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A laboratory technician drops a $0.0850$ -kg sample of unknown solid material, at $100.0$ °C, into a calorimeter. The calorimeter can, initially at $19.0$ °C, is made of $0.150$ kg of copper and contains $0.200$ kg of water. The final temperature of the calorimeter can and contents is $26.1$ °C. Compute the specific heat of the sample.

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An electric kitchen range has a total wall area of $1.40$ m² and is insulated with a layer of fiberglass $4.00$ cm thick. The inside surface of the fiberglass has a temperature of $175$ °C, and its outside surface is at $35.0$ °C. The fiberglass has a thermal conductivity of $0.040 W / m \cdot K$ . What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of $1.40$ m² ?

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A carpenter builds an exterior house wall with a layer of wood $3.0$ cm thick on the outside and a layer of Styrofoam insulation $2.2$ cm thick on the inside wall surface. The wood has $k = 0.080 W / m \cdot K$ , and the Styrofoam has $k = 0.027 W / m \cdot K$ . The interior surface temperature is $19.0$ °C, and the exterior surface temperature is $negative 10.0$ °C. What is the temperature at the plane where the wood meets the Styrofoam?

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Textbook Question

A carpenter builds an exterior house wall with a layer of wood $3.0$ cm thick on the outside and a layer of Styrofoam insulation $2.2$ cm thick on the inside wall surface. The wood has $k=0.080\,W/m$\cdot$ K$ , and the Styrofoam has $k=0.027\,W/m$\cdot$ K$ . The interior surface temperature is $19.0$ °C, and the exterior surface temperature is $-10.0$ °C. What is the rate of heat flow per square meter through this wall?

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A $4.00$ -kg silver ingot is taken from a furnace, where its temperature is $750.0$ °C, and placed on a large block of ice at $0.0$ °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 $0.250$ kg of water at $75.0$ °C. How many kilograms of ice at $negative 20.0$ °C must be dropped into the water to make the final temperature of the system $40.0$ °C?

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