Answer : The change in the internal energy of the weight lifter is, [tex]-6.24\times 10^5J[/tex]
Explanation : Given,
Mass of water = 0.200 kg
Work done = [tex]1.40\times 10^5J[/tex]
First we have to calculate the amount of heat released.
[tex]Q=-mL[/tex]
where,
Q = heat released
m = mass
L = latent heat vaporization = [tex]2.42\times 10^6J/kg[/tex]
Now put all the given values in the above formula, we get:
[tex]Q=-(0.200kg)\times (2.42\times 10^6J/kg)[/tex]
[tex]Q=-4.84\times 10^5J[/tex]
Now we have to calculate the change in the internal energy of the weight lifter.
[tex]\Delta U=q-w[/tex]
where,
[tex]\Delta U[/tex] = change in internal energy
q = heat
w = work done
Now put all the given values in the above formula, we get:
[tex]\Delta U=(-4.84\times 10^5J)-(1.40\times 10^5J)[/tex]
[tex]\Delta U=-6.24\times 10^5J[/tex]
Thus, the change in the internal energy of the weight lifter is, [tex]-6.24\times 10^5J[/tex]
