Physics College

Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy are radiation, evaporation of sweat, evaporation from the lungs, conduction, and convection. In this question, we will focus on the evaporation of sweat alone, although all of these mechanisms are needed to survive. The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).
(A) To cool the body of a jogger of mass 90 kg by 1.8°C , how much sweat has to evaporate?
O 130 g
O 230 g
O 23 g
O 13 g

Answers

Answer 1

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

     Q_lost = - Q_absorbed

     

The latent heat is

      Q_absorbed = m L

The heat given by the body

      Q_lost = M [tex]c_{e}[/tex] ΔT

       

where m is the mass of sweat and M is the mass of the body

       m L = M c_{e} ΔT

        m = M c_{e} ΔT / L

let's replace

        m = 90  3.500  1.8 / 2.42 10⁶

 

        m = 0.2343 kg

reduced to grams

        m = 0.2342 kg (1000g / 1kg)

        m = 23.42 g

 the correct answer is c) 23 g

Answer 2

230 g should have to evaporate.

Given that,

The latent heat of vaporization of sweat at body temperature (37 °C) is 2.42 x 10^6 J/kg and the specific heat of a body is approximately 3500 J/(kg*°C).

The calculation is as follows:

[tex]= \frac{(90kg)(3500J/kg^{\circ})(1.8^{\circ}C)}{2.42\times 10^6J/kg}[/tex]

= 0.23 kg

= 230g

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Related Questions

The primary coil of a transformer has N1 = 275 turns, and its secondary coil has N2 = 2,200 turns. If the input voltage across the primary coil is Δv = (160 V)sin ωt, what rms voltage is developed across the secondary coil?

Answers

Answer:

Secondary voltage of transformer is 905.23 volt  

Explanation:

It is given number of turns in primary of transformer [tex]N_1=275[/tex]

Number of turns in secondary [tex]N_2=2200[/tex]

Input voltage equation of the transformer

[tex]\Delta v=160sin\omega t[/tex]

Here [tex]v_{max}=160volt[/tex]

[tex]v_{rms}=\frac{160}{\sqrt{2}}=113.15volt[/tex]

For transformer we know that

[tex]\frac{V_1}{V_2}=\frac{N_1}{N_2}[/tex]

[tex]\frac{113.15}{V_2}=\frac{275}{2200}[/tex]

[tex]V_2=905.23Volt[/tex]

Therefore secondary voltage of transformer is 905.23 volt

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