Question

The change in e (internal energy) of a system that releases 45.0 kj of heat and is performing 855 j of work on the surroundings is ____? a. 44.1 kj b. - 44.1 kj c. -900.0 kj d. - 45.9 kj

Answer 1

The change in internal energy of the system is -45.855 kJ; when rounded to one decimal place, it is -45.9 kJ, corresponding to answer option d.

Step-by-step explanation:

To calculate the change in internal energy (ΔE) of a system, you can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings: ΔE = Q - W. In this case, 45.0 kJ of heat is released (Q = -45.0 kJ because heat is going out of the system) and the system is performing 855 J of work on the surroundings (W = 855 J because the work is done by the system).

First, we convert the work done from joules to kilojoules (since the heat is in kilojoules): 855 J = 0.855 kJ. Now we can use the formula with consistent units:

ΔE = Q - W = (-45.0 kJ) - (0.855 kJ) = -45.855 kJ

The change in internal energy of the system is -45.855 kJ, which, rounded to one decimal place, would be -45.9 kJ. Therefore, the correct answer is option d. -45.9 kJ.

Answer 2

OPTION D.

The change in internal energy of a system that releases 45.0 kJ of heat and performs 855 J of work is -45.9 kJ. This is calculated using the first law of thermodynamics, where the heat released and work done are both accounted for in determining the total change in internal energy.

Step-by-step explanation:

The change in internal energy (\( \Delta E \)) of a system in thermodynamics can be calculated using the first law of thermodynamics: \( \Delta E = Q - W \), where \(Q\) is the heat added to the system and \(W\) is the work done by the system on the surroundings. In this scenario, the system releases 45.0 kJ of heat and performs 855 J of work, which we must convert to kJ to match the units of the heat term, resulting in 0.855 kJ of work.

Now, we can substitute the values into the equation:

\( \Delta E = Q - W \)
\( \Delta E = (-45.0 \text{ kJ}) - (0.855 \text{ kJ}) \)
\( \Delta E = -45.855 \text{ kJ} \)

Thus, the change in internal energy of the system is -45.9 kJ, corresponding to option (d).