JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 11)
1 g of a liquid is converted to vapour at 3 $$\times$$ 10$$^5$$ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm$$^3$$ during this phase change, then the increase in internal energy in the process will be :
4800 J
4320 J
432000 J
4.32 $$\times$$ 10$$^8$$ J
Explanation
Work done = P$\Delta$V
= 3 × 105 × 1600 × 10–6
= 480 J
Only 10% of heat is used in work done.
Hence $\Delta$Q = 4800 J The rest goes in internal energy, which is 90% of heat.
Change in internal energy = 0.9 × 4800 = 4320 J
= 3 × 105 × 1600 × 10–6
= 480 J
Only 10% of heat is used in work done.
Hence $\Delta$Q = 4800 J The rest goes in internal energy, which is 90% of heat.
Change in internal energy = 0.9 × 4800 = 4320 J
Comments (0)
