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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 17, Problem 58a

An electric kitchen range has a total wall area of 1.401.40 m2 and is insulated with a layer of fiberglass 4.004.00 cm thick. The inside surface of the fiberglass has a temperature of 175175°C, and its outside surface is at 35.035.0°C. The fiberglass has a thermal conductivity of 0.040W/mK0.040\;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 1.401.40 m2 ?

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Identify the formula for heat current through a flat slab, which is given by Fourier's law of heat conduction: \( Q = \frac{k \cdot A \cdot (T_{inside} - T_{outside})}{d} \), where \( Q \) is the heat current, \( k \) is the thermal conductivity, \( A \) is the area, \( T_{inside} \) and \( T_{outside} \) are the temperatures of the inside and outside surfaces, and \( d \) is the thickness of the slab.
Substitute the given values into the formula: \( k = 0.040 \text{ W/m K} \), \( A = 1.40 \text{ m}^2 \), \( T_{inside} = 175°C \), \( T_{outside} = 35.0°C \), and \( d = 0.04 \text{ m} \).
Calculate the temperature difference \( \Delta T = T_{inside} - T_{outside} = 175°C - 35°C \).
Plug the values into the formula: \( Q = \frac{0.040 \cdot 1.40 \cdot (175 - 35)}{0.04} \).
Simplify the expression to find the heat current \( Q \). This will give you the rate at which heat is transferred through the fiberglass insulation.

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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 defined as the amount of heat that passes through a material with a given thickness, area, and temperature difference. In this problem, the fiberglass's thermal conductivity is crucial for calculating the heat current through the insulation.
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Heat Current

Heat current, also known as heat flow rate, is the rate at which heat energy transfers through a material. It is calculated using the formula Q = kA(T1 - T2)/d, where k is the thermal conductivity, A is the area, T1 and T2 are the temperatures on either side of the material, and d is the thickness. Understanding this concept is essential for solving the problem of heat transfer through the fiberglass.
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Temperature Gradient

The temperature gradient is the change in temperature across a material per unit distance. It is a driving factor for heat transfer, as heat flows from regions of higher temperature to lower temperature. In this scenario, the temperature difference between the inside and outside surfaces of the fiberglass creates a gradient that influences the heat current calculation.
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Related Practice
Textbook Question

The emissivity of tungsten is 0.3500.350. A tungsten sphere with radius 1.501.50 cm is suspended within a large evacuated enclosure whose walls are at 290.0290.0 K. What power input is required to maintain the sphere at 3000.03000.0 K if heat conduction along the supports is ignored?

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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 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

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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