Math / Calculus

QuestionProblem 5: (13\% of Assignment Value) Suppose a human body's core internal temperature is $37.0^{\circ} \mathrm{C}$ , the skin temperature is $34.0^{\circ} \mathrm{C}$ , the thickness of the tissues between averages 1.2 cm , and the surface area is $1.33 \mathrm{~m}^{2}$ . Martin, Charlie - charlierob7777@outlook com (a) theexpertia com - tracking id:7M79-7F-79-40-93B4-44857: In accordance with Expert TA's ferms of Service. copying this . information to any solutions sharing tebsite is strictly forbidden. Doing so may fesult intermination of jourt Expert TAA Account. \begin{tabular}{l|l} \multicolumn{1}{c|}{ Substance } & \\ \hline Glass wool & \\ \hline Wool & \\ \hline Glass & \\ \hline Human body & \\ \hline Ceramic & \\ \hline Wood & \\ \hline Air & \\ \hline Fatty tissue & \\ \hline Styrofoam & \end{tabular}
Calculate the rate of heat conduction, in watts, out of the human body. $Q / \Delta t=$ $\square$ $\sin ()$ $\cos ()$ $\tan ()$ 7 8 9 HOME $\operatorname{cotan}()$ asin() $\operatorname{acos}()$ 4 5 6 $\square$ $\operatorname{atan}()$ $\operatorname{acotan}$ () $\sinh ()$ 1 2 3 0 $\operatorname{cotanh}()$ $\square$ END

Studdy Solution

STEP 1

What is this asking? How much heat, in watts, is flowing out of a person's body, given their internal and skin temperatures, the thickness of their tissues, and their surface area? Watch out! Make sure to use the right thermal conductivity for human tissue, convert everything to consistent units, and remember that watts are joules per second.

STEP 2

1. Find the thermal conductivity.
2. Calculate the temperature difference.
3. Compute the heat flow rate.

STEP 3

We need the thermal conductivity of human tissue.
A quick search tells us it's roughly $0.2 \, \text{W/m}\cdot\text{K}$ .
Remember, this tells us how easily heat flows through the material.
A higher value means heat flows more easily.

STEP 4

The temperature difference between the core and skin is what drives the heat flow.
We have $37.0^\circ \text{C}$ inside and $34.0^\circ \text{C}$ on the skin.
So the difference is $37.0 - 34.0 = \mathbf{3.0^\circ \text{C}}$ .
Since we're dealing with a temperature difference, this is the same as $\mathbf{3.0 \, \text{K}}$ .
Remember, a change of one degree Celsius is the same as a change of one Kelvin!

STEP 5

Now, we use the formula for heat conduction: $\frac{Q}{\Delta t} = k \cdot A \cdot \frac{\Delta T}{d}$ where $\frac{Q}{\Delta t}$ is the rate of heat flow (what we want!), $k$ is the thermal conductivity, $A$ is the surface area, $\Delta T$ is the temperature difference, and $d$ is the tissue thickness.

STEP 6

Let's plug in our values.
We have $k = \mathbf{0.2 \, \text{W/m}\cdot\text{K}}$ , $A = \mathbf{1.33 \, \text{m}^2}$ , $\Delta T = \mathbf{3.0 \, \text{K}}$ , and $d = \mathbf{1.2 \, \text{cm}} = \mathbf{0.012 \, \text{m}}$ (don't forget to convert centimeters to meters!).

STEP 7

So, we get: $\frac{Q}{\Delta t} = (0.2 \, \text{W/m}\cdot\text{K}) \cdot (1.33 \, \text{m}^2) \cdot \frac{3.0 \, \text{K}}{0.012 \, \text{m}}$

STEP 8

Notice how the units cancel nicely, leaving us with watts: $\frac{Q}{\Delta t} = \frac{0.2 \cdot 1.33 \cdot 3.0}{0.012} \, \text{W}$

STEP 9

Calculating this gives us: $\frac{Q}{\Delta t} = \mathbf{66.5 \, \text{W}}$

STEP 10

The rate of heat conduction out of the human body is $\mathbf{66.5 \, \text{W}}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the thermal conductivity.
2. Calculate the temperature difference.
3. Compute the heat flow rate.

STEP 3

We need the thermal conductivity of human tissue.
A quick search tells us it's roughly $0.2 \, \text{W/m}\cdot\text{K}$ .
Remember, this tells us how easily heat flows through the material.
A higher value means heat flows more easily.

STEP 4

STEP 5

STEP 6

Let's plug in our values.
We have $k = \mathbf{0.2 \, \text{W/m}\cdot\text{K}}$ , $A = \mathbf{1.33 \, \text{m}^2}$ , $\Delta T = \mathbf{3.0 \, \text{K}}$ , and $d = \mathbf{1.2 \, \text{cm}} = \mathbf{0.012 \, \text{m}}$ (don't forget to convert centimeters to meters!).

STEP 7

So, we get: $\frac{Q}{\Delta t} = (0.2 \, \text{W/m}\cdot\text{K}) \cdot (1.33 \, \text{m}^2) \cdot \frac{3.0 \, \text{K}}{0.012 \, \text{m}}$

STEP 8

Notice how the units cancel nicely, leaving us with watts: $\frac{Q}{\Delta t} = \frac{0.2 \cdot 1.33 \cdot 3.0}{0.012} \, \text{W}$

STEP 9

Calculating this gives us: $\frac{Q}{\Delta t} = \mathbf{66.5 \, \text{W}}$

STEP 10

The rate of heat conduction out of the human body is $\mathbf{66.5 \, \text{W}}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the thermal conductivity.2. Calculate the temperature difference.3. Compute the heat flow rate.

STEP 3

We need the thermal conductivity of human tissue.A quick search tells us it's roughly 0.2 W/m⋅K0.2 \, \text{W/m}\cdot\text{K}0.2W/m⋅K.Remember, this tells us how easily heat flows through the material.A higher value means heat flows more easily.

STEP 4

STEP 5

STEP 6

STEP 7

So, we get: QΔt=(0.2 W/m⋅K)⋅(1.33 m2)⋅3.0 K0.012 m \frac{Q}{\Delta t} = (0.2 \, \text{W/m}\cdot\text{K}) \cdot (1.33 \, \text{m}^2) \cdot \frac{3.0 \, \text{K}}{0.012 \, \text{m}} ΔtQ​=(0.2W/m⋅K)⋅(1.33m2)⋅0.012m3.0K​

STEP 8

Notice how the units cancel nicely, leaving us with watts: QΔt=0.2⋅1.33⋅3.00.012 W \frac{Q}{\Delta t} = \frac{0.2 \cdot 1.33 \cdot 3.0}{0.012} \, \text{W} ΔtQ​=0.0120.2⋅1.33⋅3.0​W

STEP 9

Calculating this gives us: QΔt=66.5 W \frac{Q}{\Delta t} = \mathbf{66.5 \, \text{W}} ΔtQ​=66.5W

STEP 10

The rate of heat conduction out of the human body is 66.5 W\mathbf{66.5 \, \text{W}}66.5W.

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1. Find the thermal conductivity.
2. Calculate the temperature difference.
3. Compute the heat flow rate.

We need the thermal conductivity of human tissue.
A quick search tells us it's roughly $0.2 \, \text{W/m}\cdot\text{K}$ .
Remember, this tells us how easily heat flows through the material.
A higher value means heat flows more easily.

So, we get: $\frac{Q}{\Delta t} = (0.2 \, \text{W/m}\cdot\text{K}) \cdot (1.33 \, \text{m}^2) \cdot \frac{3.0 \, \text{K}}{0.012 \, \text{m}}$

Notice how the units cancel nicely, leaving us with watts: $\frac{Q}{\Delta t} = \frac{0.2 \cdot 1.33 \cdot 3.0}{0.012} \, \text{W}$

Calculating this gives us: $\frac{Q}{\Delta t} = \mathbf{66.5 \, \text{W}}$

The rate of heat conduction out of the human body is $\mathbf{66.5 \, \text{W}}$ .