A rapidly spinning paddle wheel raises the temperature of 200mL of water from 21 degrees Celsius to 25 degrees. How much
the work is done by the friction force. All the work will go to increase the internal energy of the water.
Q=mC∆θ=3360 J
Where C=4200J, m=1000×200×10^-6=0.2 kg
There are ”No” heat transfer in the process.
W = Ht = .200(25 – 21)4184 = 3347 J
specific heat of water is 4.186 kJ/kgC
200 mL = 0.2 kg
E = 4.186 kJ/kgC x 0.2 kg x (25–21)C = 3.35 kJ
which is the work done, and the heat (thermal energy) transferred
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