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In exercising, a weight lifter loses 0.178 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.35E+5 J. Assuming that the latent heat of vaporization of perspiration is 2.44E+6 J/kg, find the change in the internal energy of the weight lifter.

Sagot :

The change in the internal energy of the weight lifter is -5.69 x 10⁵J.

Solution:

[tex]\Delta U = Q-W[/tex]

work done is lifting a body is [tex]1.35*10^5 J[/tex]

so, [tex]W = 1.35*10^5 J[/tex]

during perspiration, heat will be lost by the body to make water evaporate

so,

[tex]Q= -L_v \times m[/tex]

where m is the mass of water

[tex]Q= -2.44 \times 10^6 \times 0.178 J=-4.34 \times 10^5 J[/tex]

Now, change in internal energy,

[tex]\Delta U = -4.34\times 10^5- 1.35 \times 10^5 =-5.69 \times 10^5 J[/tex]

A glass of water standing on a table has no visible macroscopic potential or kinetic energy. However, water molecules have internal energy because they move randomly and each molecule contributes to the total internal energy of water. The energy involved in the system is related to the random motion of the particles and the potential energy of the molecules due to their orientation.

Learn more about Internal energy here:- https://brainly.com/question/25737117

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