Skip to main content
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 59b

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?

Verified step by step guidance
1
First, understand that the problem involves calculating the rate of heat flow through a composite wall made of two materials: wood and Styrofoam. This requires using the concept of thermal resistance and the formula for heat transfer through a layered wall.
The formula for the rate of heat flow (Q) through a wall is given by: ΔTRtotal, where ΔT is the temperature difference across the wall, and Rtotal is the total thermal resistance of the wall.
Calculate the temperature difference ΔT between the interior and exterior surfaces: ΔT = 19.0 - - 10.0 = 29.0 °C.
Calculate the thermal resistance for each layer using the formula: Ri = diki, where di is the thickness of the layer and ki is the thermal conductivity. Calculate Rwood and Rstyrofoam.
Sum the thermal resistances to find the total resistance: Rtotal = Rwood + Rstyrofoam. Finally, use the formula for heat flow to find the rate of heat flow per square meter: ΔTRtotal.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
17m
Was this helpful?

Key Concepts

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

Thermal Conductivity

Thermal conductivity (k) is a material property that indicates how well a material conducts heat. It is measured in watts per meter per degree Kelvin (W/m K). Materials with high thermal conductivity transfer heat quickly, while those with low thermal conductivity are better insulators. In this problem, wood and Styrofoam have different thermal conductivities, affecting the rate of heat flow through the wall.
Recommended video:

Heat Transfer Through Composite Walls

Heat transfer through composite walls involves calculating the rate of heat flow through multiple layers of different materials. Each layer has its own thickness and thermal conductivity, which together determine the overall resistance to heat flow. The heat flow rate can be calculated using the formula for thermal resistance, considering the temperature difference across the wall and the properties of each layer.
Recommended video:
Guided course
05:14
Overview of Heat Transfer

Temperature Gradient

The temperature gradient is the change in temperature across a material, which drives heat flow. In this scenario, the temperature difference between the interior (19.0°C) and exterior (-10.0°C) surfaces creates a gradient that causes heat to flow from the warmer interior to the cooler exterior. Understanding this gradient is crucial for calculating the rate of heat transfer through the wall.
Recommended video:
Guided course
04:54
Introduction To Temperature Scales
Related Practice
Textbook Question

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 37.037.0°C and that the outer surface of the skin is at 30.030.0°C. Skin varies in thickness from 0.500.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.750.75 mm. A 165165-lb, 66-ft-tall person has a surface area of about 2.02.0 m2 and loses heat at a net rate of 7575 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person's skin?

2
views
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?

1
views
Textbook Question

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 ?

1
views
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?

2
views
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?

1
views
Textbook Question

A spherical pot contains 0.750.75 L of hot coffee (essentially water) at an initial temperature of 9595°C. The pot has an emissivity of 0.600.60, and the surroundings are at 20.0 20.0°C. Calculate the coffee's rate of heat loss by radiation.

2
views