Question

A moving car has an internal energy of 2,769 J. Pressing the accelerator adds 527 J of energy to the car. Thesystem now has an internal energy of 3,228 J. What isthe DeltaQ? Give exact value.E+=B+DeltaW+deltaQ

Answer 1

Given data:

* The initial internal energy of the car is E_i = 2769 J.

* The amount of work done on the car to accelerate is dW = 527 J.

* The final internal energy of the car is E_f = 3228 J.

Solution:

From the first law of thermodynamics, the heat energy loss by the car is,

$\begin{gathered} E_f-E_i=dW-(-dQ) \\ E_f-E_i=dW+dQ \end{gathered}$

Where the negative sign indicates the loss of heat energy from the car,

Substituting the known values,

$\begin{gathered} 3228-2769=527+dQ \\ 459=527+dQ \\ dQ=459-527 \\ dQ=-68\text{ J} \end{gathered}$

Thus, the amount of energy loss by the car in form of heat is -68 J.