Question

A closed system undergoes a process resulting in a work output of 150 J while the heat removed from the system is 100 J. Calculate the change in internal energy of the system.

• A. -50 J
• B. 50 J
• C. -250 J
• D. 250 J

Hint:

First Law of Thermodynamics. \(\Delta U = Q - W\). Q is positive for heat added TO the system, negative for heat removed FROM the system. W is positive for work done BY the system, negative for work done ON the system. Be careful with signs.

Answer 1

The Correct Option is C

We use the First Law of Thermodynamics for a closed system: $$ \Delta U = Q - W $$ where: \(\Delta U\) = change in internal energy \(Q\) = heat added *to* the system \(W\) = work done *by* the system Given: Work output *by* the system = 150 J \(\implies W = +150\) J.
Heat *removed from* the system = 100 J \(\implies\) Heat added *to* the system \(Q = -100\) J.
Substitute these values into the first law: $$ \Delta U = (-100 \, \text{J}) - (+150 \, \text{J}) $$ $$ \Delta U = -100 - 150 $$ $$ \Delta U = -250 \, \text{J} $$ The change in internal energy of the system is -250 J.
(Note: This contradicts the user-provided key for Q80, which was 2 = 50J.
The calculation based on standard sign conventions consistently gives -250J).