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?

Ch 17: Temperature and Heat

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

All textbooks

Young & Freedman Calc 14th Edition

Ch 17: Temperature and Heat

Problem 61

Chapter 17, Problem 61

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 $37.0$ °C and that the outer surface of the skin is at $30.0$ °C. Skin varies in thickness from $0.50$ mm to a few millimeters on the palms and soles, so assume an average thickness of $0.75$ mm. A $165$ -lb, $6$ -ft-tall person has a surface area of about $2.0$ m² and loses heat at a net rate of $75$ W while resting. On the basis of our assumptions, what is the thermal conductivity of this person's skin?

Verified step by step guidance

Start by understanding the concept of thermal conductivity, which is a measure of a material's ability to conduct heat. It is denoted by the symbol 'k' and is typically measured in watts per meter per degree Celsius (W/m·°C).

Use the formula for heat transfer through conduction: $ Q = \frac{k \, A \, \Delta T}{d} $, where $ Q $ is the heat transfer rate (75 W in this case), $ A $ is the surface area (2.0 m²), $ \Delta T $ is the temperature difference (37.0°C - 30.0°C), and $ d $ is the thickness of the skin (0.75 mm, converted to meters).

Convert the thickness of the skin from millimeters to meters by dividing by 1000, so $ d = 0.75 \text{ mm} = 0.00075 \text{ m} $.

Substitute the known values into the formula: $ 75 = \frac{k \, \times \, 2.0 \, \times \, (37.0 - 30.0)}{0.00075} $.

Solve for $ k $ by rearranging the equation: $ k = \frac{75 \, \times \, 0.00075}{2.0 \, \times \, 7.0} $. This will give you the thermal conductivity of the skin in W/m·°C.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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 heat is conducted through a material. It is defined as the amount of heat that passes through a unit area of a material in a unit time for a unit temperature gradient. In this context, it helps determine how efficiently heat is transferred from the blood through the skin to the surface.

Heat Transfer

Heat transfer is the process of thermal energy moving from a hotter object to a cooler one. It can occur through conduction, convection, or radiation. In this scenario, conduction is the primary mode of heat transfer, as heat moves from the blood at the bottom layer of the skin to the outer surface, driven by the temperature difference.

Fourier's Law of Heat Conduction

Fourier's Law states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the area through which the heat flows. It is expressed as Q = -kA(dT/dx), where Q is the heat transfer rate, k is the thermal conductivity, A is the area, and dT/dx is the temperature gradient. This law is essential for calculating the thermal conductivity of the skin in the given problem.

Recommended video:

Guided course

10:01

Conduction

Related Practice

Textbook Question

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 ?

views

Textbook Question

The emissivity of tungsten is $0.350$ . A tungsten sphere with radius $1.50$ cm is suspended within a large evacuated enclosure whose walls are at $290.0$ K. What power input is required to maintain the sphere at $3000.0$ K if heat conduction along the supports is ignored?

views

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 $negative 10.0$ °C. What is the temperature at the plane where the wood meets the Styrofoam?

views

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?

views

Textbook Question

A spherical pot contains $0.75$ L of hot coffee (essentially water) at an initial temperature of $95$ °C. The pot has an emissivity of $0.60$ , and the surroundings are at $20.0$ °C. Calculate the coffee's rate of heat loss by radiation.

views