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In moving out of a dormitory at the end of the semester, a student does 2.17 x 104 J of work. In the process, his internal energy decreases by 5.03 x 104 J. Determine each of the following quantities (including the algebraic sign): (a)W, (b) U, and (c)Q.

Respuesta :

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Answer:

a. W = 2.17 x 10 ⁴ J

b. ΔU = - 5.03 x 10⁴ J

c. Q = 2.86 x 10 ⁻³ J

Explanation:

a.

The student work does 2.17 x 10⁴ J

So W = 2.17 x 10 ⁴ J  

b.

The internal energy decreases 5.03 x 10⁴ J

ΔU = U₁ - U₂

ΔU = 0 - 5.03 x 10⁴ J = - 5.03 x 10⁴ J

c.

The formula heat can use:

Q = W + ΔU

Q = 2.17 x 10⁴ J + ( - 5.03 x 10⁴ J )

Q = 2.86 x 10 ⁻³ J