Suppose you warm up 510 grams of water (about half a liter, or about a pint) on

Question

Suppose you warm up 510 grams of water (about half a liter, or about a pint) on a stove, and while this is happening, you also stir the water with a beater, doing 3 times 10^4 J of work on the water. After the large-scale motion of the water has dissipated away, the temperature of the water is observed to have risen from 19 degree C to 84 degree C. What was the change in the thermal energy of the water? Delta E_thermal = J Taking the water as the system, how much energy transfer due to a temperature difference (microscopic work) Q was there across the system boundary? Q = J Taking the water as the system, what was the energy change of the surroundings? Delta E_surroundings = J

Explanation / Answer

Ethermal = Work done on water = 3*10^4 J

Q=mcT = 0.520*4181*(84-19)= 141318 J

Esurrounding = Ethermal –Q = (3*10^4)- 141318 = -111318 J …work is done by the system

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