A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

Short Answer

Expert verified
The change in internal energy (ΔE) of the system is \(-229 \mathrm{~kJ}\).

Step by step solution

01

Identify the given values

The problem states that the system releases 125 kJ of heat (Q): that means, Q = -125 kJ (since heat is released, we consider it as negative). Furthermore, 104 kJ of work is done on the system (W): this implies, W = +104 kJ (since work is done on the system, we consider it as positive).
02

Apply the first law of thermodynamics

Now, we will use the first law of thermodynamics to find the change in the internal energy of the system (ΔE). The formula is: ΔE = Q - W.
03

Substitute the given values

We will plug in the given values of Q and W into the formula: ΔE = (-125 kJ) - (+104 kJ)
04

Calculate the change in internal energy (ΔE)

Now, perform the calculation: ΔE = -125 kJ - 104 kJ = -229 kJ The change in internal energy (ΔE) of the system is -229 kJ. Since the result is negative, it indicates that the internal energy of the system has decreased.

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