Question

Calculate the change internal energy (δe) for a system that is giving off 45.0 kj of heat and is performing 855 j of work on the surroundings.

Answer 1

The change in internal energy of a system is given by (second law of thermodynamics)
$\Delta U = Q + W$
where Q is the heat absorbed by the system and W is the work done on the system.

In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings

In our problem, the heat released by the system is $Q=-45 kJ=-45000 J$ (with negative sign since it is released by the system), and the work done is $W=-855 J$ still with negative sign because it is performed by the system on the surrounding, so the change in internal energy is
$\Delta U = Q +W=-45000 J - 855 J=-45855 J$

Answer 2

Change in internal energy is -45855 J.

Given :

Heat, Q = -45 kJ

(Q is negative because heat is released by the system)

Work, W = -855 J

(W is negative because work is done by the system)

Solution :

We know that the change in internal energy is given by,

$\rm \bigtriangleup U = Q+W$

$\rm \bigtriangleup U = -45000-855$

$\rm \bigtriangleup U = -45855\;J$

Therefore change in internal energy is -45855 J.

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https://brainly.com/question/20920764?referrer=searchResults