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Calculate \(\delta E\) if a system releases 66.9 kJ of heat to its surroundings while the surroundings do 46.0 kJ of work on the system.

Answer :

Final answer:

The change in internal energy (δe) of the system, when it releases 66.9 kJ of heat to the surroundings and 46.0 kJ of work is done on it, is calculated using the formula δe = q - w. The result is δe = -20.9 kJ.

Explanation:

To calculate the change in internal energy (δe) of a system when it releases heat and work is done on it, we can use the first law of thermodynamics. This law states that the change in internal energy of a system is equal to the heat added to the system (q) minus the work done by the system (w). In formula terms, δe = q - w.

In this scenario, the system releases 66.9 kJ of heat, which means q = -66.9 kJ because the heat is being lost to the surroundings. The surroundings do 46.0 kJ of work on the system, which means w = 46.0 kJ. Remember that when work is done on the system, it's considered positive from the system's perspective.

δe = q - w = -66.9 kJ - (-46.0 kJ) = -66.9 kJ + 46.0 kJ = -20.9 kJ. Therefore, the change in internal energy of the system is -20.9 kJ.

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Rewritten by : Barada