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Otrada [13]
2 years ago
14

Assume that the thickness of the tissue between the interior and the exterior of the body is =3 cm and that the average area thr

ough which conduction can occur is =1.5 m2 . Find heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2 0C. Coefficient of thermal for tissue without blood is = 18 × 10−2 Cal /m-hr- 0C.
Physics
1 answer:
Westkost [7]2 years ago
7 0

The heat transferred by conduction per hour if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.

<h3>What is heat conduction?</h3>

Heat conduction is the method of transfer of heat by the direct contact of the two or more solid bodies.

Given is the thickness of tissue x = 3cm =-0.03m, Area of conduction A = 1.5 m², Change in temperature of skin ΔT =2°C, and coefficient of thermal for tissue without blood is k = 18 × 10⁻² Cal /m-hr- °C

The conduction heat rate in hr is

Q =kAΔT /x

Q =18 × 10⁻² x 1.5 x 2 /0.03

Q =18 Joule / hr

Thus, the heat transferred by conduction per hour, if temperature difference between the inner body and the skin is ∆ = 2°C is 18 Joule/ hr.

Learn more about conduction.

brainly.com/question/12136944

#SPJ1

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Answer:

c = 0.4356 J/gK

Explanation:

Given the following data;

Mass = 450 grams

Initial temperature, T1 = 150°C

Final temperature, T2 = 53°C

Quantity of heat = 34500 Joules

To find the specific heat capacity of the metal;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 53 - 150

dt = -97°C

Converting the temperature in Celsius to Kelvin, we have;

dt = 273 + (-97) = 176 Kelvin

Making c the subject of formula, we have;

c = \frac {Q}{mdt}

Substituting into the equation, we have;

c = \frac {34500}{450*176}

c = \frac {34500}{79200}

c = 0.4356 J/gK

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A wire 1.0 m long experiences a magnetic force of 0.50 N due to a
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3 years ago
In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

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Hope this helps! I would greatly appreciate a brainliest! :)
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