$2 \cdot 5 \mathrm{~kJ}$ of work is done on the system and it releases $1500 \mathrm{~J}$ of heat. What is…
$2 \cdot 5 \mathrm{~kJ}$ of work is done on the system and it releases $1500 \mathrm{~J}$ of heat. What is the change in internal energy?
- $1000 \mathrm{~J}$
- $4000 \mathrm{~J}$
- $2500 \mathrm{~J}$
- $1500 \mathrm{~J}$
Solution
$\Delta U=q+w = -1500 + 2500 = 1000 ~J$
Asked in: MHT CET 2020 (20 Oct Shift 2)
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