Question

in a process a system does 180j of work on the surroundings and only 40j of heat is added to the system hence change in internal energy is

Answer 1

-140 J

Answer 2

The first law of thermodynamics relates the change in internal energy ($\Delta U$) of a system to the heat ($Q$) added to the system and the work ($W$) done on the system. The equation is:

$\Delta U = Q + W$

We need to use the standard sign conventions:

• Heat ($Q$):
  • Heat added to the system is positive ($Q > 0$).
  • Heat removed from the system is negative ($Q < 0$).
• Work ($W$):
  • Work done on the system is positive ($W > 0$).
  • Work done by the system is negative ($W < 0$).

From the problem statement:

1. "a system does 180 J of work on the surroundings"
   This means work is done by the system. According to our convention, $W = -180 \, \text{J}$.
2. "only 40 J of heat is added to the system"
   This means heat is added to the system. According to our convention, $Q = +40 \, \text{J}$.

Now, substitute these values into the first law of thermodynamics:

$\Delta U = Q + W$ $\Delta U = (+40 \, \text{J}) + (-180 \, \text{J})$ $\Delta U = 40 \, \text{J} - 180 \, \text{J}$ $\Delta U = -140 \, \text{J}$

The change in internal energy is -140 J.