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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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A 259-kV power transmission line carrying 429 A is hung from grounded metal towers by ceramic insulators, each having a 0.71×109
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Answer:

The resistance is 7.47\times10^{6}\ \Omega.

Explanation:

Given that,

Power = 259 kV

Current = 429 A

Resistance R=0.71\times10^{9}\ Omega

We need to calculate the current in each insulator

Using formula of current

I=\dfrac{P}{R}

Put the value into the formula

I=\dfrac{259\times10^{3}}{0.71\times10^{9}}

I=3.64\times10^{-4}\ A

So all 95 insulators are in parallel

We need to calculate the resistance

Using formula of resistance

\dfrac{1}{R}=\sum_{i=1}^{95}\dfrac{1}{R_{i}}

Put the value into the formula

\dfrac{1}{R}=\dfrac{95}{0.71\times10^{9}}

\dfrac{1}{R}=1.338\times10^{-7}

R=\dfrac{1}{1.338\times10^{-7}}

R=7473841.5=7.47\times10^{6}\ \Omega

Hence, The resistance is 7.47\times10^{6}\ \Omega.

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4 years ago
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Korvikt [17]

Answer:

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Explanation:

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Answer:

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