Suppose you warm up 470 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 4104 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 25â°c to 76â°c. what was the change in the thermal energy of the water?

104,442 J

Explanation:

First of all, we need to calculate the amount of heat Q supplied to the water, given by:

where

m = 470 g is the mass of the water

is the specific heat capacity of the water

is the change in temperature

Substituting,

Now we can calculate the change in thermal energy of the water, , by using the 1st law of thermodynamics:

where W = 4104 J is the work done on the water. Substituting,

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