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worty [1.4K]
3 years ago
14

On a cold winter day when the temperature is −20∘C, what amount of heat is needed to warm to body temperature (37 ∘C) the 0.50 L

of air exchanged with each breath? Assume that the specific heat of air is 1020 J/kg⋅K and that 1.0 L of air has mass 1.3×10−3kg.
Physics
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

75.6J

Explanation:

Hi!

To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,

Given the above we have the following equation.

Q=(m)(h2)-(m)(h1)

where

m=mass=1.3×10−3kg.

h2= entalpy at 37C

h1= entalpy at -20C

Q=m(h2-h1)

remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference

Q=mCp(T2-T1)

Cp= specific heat of air = 1020 J/kg⋅K

Q=(1.3×10−3)(1020)(37-(-20))=75.6J

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Answer:

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Explanation:

Let's calculate the work and the magnetic force, the expression for magnetic force is

        F = qv x B

Bold indicate vector quantities, the expression for the job is

         W = F. X

Let's replace in this equation

       W = q v x B . X

The definition of speed is

      v =  dX / dt

With what work is left

     W = q dX / dt x B . X

As we can see the vector product gives us a vector perpendicular to dX and its scalar product by X of zero

Second part

   The speed a vector and although the magnitude is constant the change of direction implies a change in the speed.

   Let's calculate the magnitudes of speed (speed)

       F = qv B sin θ

       F = ma

       q v B sin θ = ma

       a = qvB / m senT

     

This acceleration is perpendicular to the magnetic field and the velocity, so it does not change if magnitude but its direction, it is directed to the center of the circle.

   | v | = q vB/m sin θ

3 0
4 years ago
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Answer:

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Answer:

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Explanation:

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