JEE Main 2019 — Chemistry solution

JEE Main2019Chemistry

During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, $Δ$ U (in kJ) is.

Solution

$w = 10 k J q = - 2 k J Δ U = q + w = 10 - 2 = 8 k J$

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