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Suppose you warm up 550 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 4e4 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 24°C to 75°C.

What was the change in the thermal energy of the water?

The answer is 1.18e5 Joules

The equation is Delta E = W + Q = mC*(Delta Temp)

Solve for Q m=550 C=4.7 change in temp=51 and I'm guessing work is 4e4, but solving for Q using those numbers gives me 2.945 Joules. What am I messing up?

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# Thermal Engergy and Water

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