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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 25°c to 76°c. what was the change in the thermal energy of the water?

Respuesta :

skyluke89

Answer:

104,442 J

Explanation:

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

[tex]Q=mC_s \Delta T[/tex]

where

m = 470 g is the mass of the water

[tex]C_s = 4.186 J/gC[/tex] is the specific heat capacity of the water

[tex]\Delta T=76 C-25 C=51 C[/tex] is the change in temperature

Substituting,

[tex]Q=(470 g)(4.186 J/gC) (51 C)=100,338 J[/tex]

Now we can calculate the change in thermal energy of the water, [tex]\Delta U[/tex], by using the 1st law of thermodynamics:

[tex]\Delta U=Q+W[/tex]

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

[tex]\Delta U=100,338 J+4,104 J=104,442 J[/tex]

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