Search any question & find its solution
Question:
Answered & Verified by Expert
The internal energy change in a system that has absorbed $2 \mathrm{kcal}$ of heat and done $500 \mathrm{~J}$ of work is
Options:
Solution:
2833 Upvotes
Verified Answer
The correct answer is:
$7900 \mathrm{~J}$
Key Idea Heat given to a system ( $\Delta Q$ ) is equal to the sum of increase in the internal energy $(\Delta \mathrm{u})$ and the work done $(\Delta W)$ by the system against the surrounding and $1 \mathrm{cal}=4.2 \mathrm{~J}$.
According to first law of thermodynamics
$$
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{Q}-\mathrm{W} \\
& =2 \times 4.2 \times 1000-500 \\
& =8400-500 \\
& =7900 \mathrm{~J}
\end{aligned}
$$
According to first law of thermodynamics
$$
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{Q}-\mathrm{W} \\
& =2 \times 4.2 \times 1000-500 \\
& =8400-500 \\
& =7900 \mathrm{~J}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.