Question

Work done by the system is $300$ J when $100$ calories heat is supplied to it. The change in internal energy (in Joules) during the process is:

• A. $-200$
• B. $+400$
• C. $+720$
• D. $+120$

Answer 1

Answer: (D)

$+120$

Change in internal energy $=$ heat supplied $+$ work done

The formula is given below:

$\Delta U=q+w$

As per sign convention, work done by the system on the surrounding has a negative sign and the heat absorbed by the system from the surrounding has a positive sign.

Hence, $w=-300$ J

and $q=100$ cal $=420.0$ J ($\because 1$ cal $=4.2$ J)

Substituting values in the expression for the change in internal energy, we get

$\Delta U=q+w=420-300=120$ J

The change in internal energy during the process is $+120$ J.

Hence, the correct option is $D$