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 21degreeC to 78degreeC What was the change in the thermal energy of the water? DeltaEthermal = 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? Taking the water as the system, what was the energy change of the surroundings? DeltaEsurroundings = j

Explanation / Answer

Here ,

work done on water , W = 3 *10^4 J

change in thermal energy of water = m * S * delta T

change in thermal energy of water = 510 *4.186 * (78 -21)

change in thermal energy of water =121687 J

the change in thermal energy of water is 121687 J

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heat = change in internal energy -work done

heat = 121687 -3 *10^4

heat = 91687 J

heat is 91687 J

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energy change of surronding = -energy change of water

energy change of surronding = -121687 J

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