
The normal body-temperature of a person is \[97^\circ F\] calculate the rate at which heat is flowing out of his body through the clothes assuming the following values.Room temperature $ = 47^\circ F$ ,surface of body under clothes $ = 1.6{m^2}$ conductivity of cloth is $0.04\,J{s^{ - 1}}{m^{ - 1}}^\circ {C^{ - 1}}$ ,thickness of cloth is $ = 0.5\,cm$.
Answer
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Hint: In order to solve this question, we will understand the definition of heat flow which states that heat flow is energy that flows from one side to another due to difference in temperature. It can flow through three different methods, conduction, convection and radiation. Conduction does not involve movement of mass whereas convection does not. Heat flows out through body through radiation which does not require any material medium to transfer heat
Complete step by step answer:
To calculate heat flow through body we first need to convert $^\circ F \to ^\circ C$ using relation,
$^\circ C = \dfrac{{(^\circ F - 32)}}{{1.8}}$
According to the given parameters in question we have,
Here, ${T_1} = 94^\circ F$ changes using relation.
${T_1}' = \dfrac{{(94 - 32)}}{{1.8}} = 34^\circ C$
And ${T_2} = 47^\circ F$ changes using relation.
${T_2}' = \dfrac{{(47 - 32)}}{{1.8}} = 8.34^\circ C$
Now by using relation $Q = (\dfrac{{ka}}{l})({T_1}' - {T_2}')$
Where, $Q$ is the rate of heat flow out from the body, $k$ is conductivity and $a$ is area of cloth.
Given in question $k = 0.04J{s^{ - 1}}{m^{ - 1}}^\circ {C^{ - 1}}$ $a = 1.6{m^2}$ and $l = 0.5cm = 0.005m$
Putting these values in equation we get
$Q = 0.04 \times (\dfrac{{1.6}}{{0.005}}) \times (34 - 8.34)$
$ \therefore Q = 328.44\,J{s^{ - 1}}$
So the answer is $328J{s^{ - 1}}$ rate at which energy is flowing outside the body.
Note: It should be remembered that we need to convert from Fahrenheit to Celsius because of the unit of conductivity so that we can get the rate of heat at which energy is flowing out through the body. Also it should be remembered that the radiation mode of heat transfer at many places involves photons of energy and conduction involves phonons to transfer heat but convection transfers heat through transfer of actual matter.
Complete step by step answer:
To calculate heat flow through body we first need to convert $^\circ F \to ^\circ C$ using relation,
$^\circ C = \dfrac{{(^\circ F - 32)}}{{1.8}}$
According to the given parameters in question we have,
Here, ${T_1} = 94^\circ F$ changes using relation.
${T_1}' = \dfrac{{(94 - 32)}}{{1.8}} = 34^\circ C$
And ${T_2} = 47^\circ F$ changes using relation.
${T_2}' = \dfrac{{(47 - 32)}}{{1.8}} = 8.34^\circ C$
Now by using relation $Q = (\dfrac{{ka}}{l})({T_1}' - {T_2}')$
Where, $Q$ is the rate of heat flow out from the body, $k$ is conductivity and $a$ is area of cloth.
Given in question $k = 0.04J{s^{ - 1}}{m^{ - 1}}^\circ {C^{ - 1}}$ $a = 1.6{m^2}$ and $l = 0.5cm = 0.005m$
Putting these values in equation we get
$Q = 0.04 \times (\dfrac{{1.6}}{{0.005}}) \times (34 - 8.34)$
$ \therefore Q = 328.44\,J{s^{ - 1}}$
So the answer is $328J{s^{ - 1}}$ rate at which energy is flowing outside the body.
Note: It should be remembered that we need to convert from Fahrenheit to Celsius because of the unit of conductivity so that we can get the rate of heat at which energy is flowing out through the body. Also it should be remembered that the radiation mode of heat transfer at many places involves photons of energy and conduction involves phonons to transfer heat but convection transfers heat through transfer of actual matter.
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