If a system has 2.50 × 102 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in internal energy of the system?

The change in internal energy of a system is given by the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W): ΔU = Q - W.

In this problem, the work is done on the system, so the work term is positive. Also, the heat is released by the system into its surroundings, so the heat term is negative.

First, we need to convert the given values into the same unit, since the work is given in kcal and the heat is given in kJ. We know that 1 kcal = 4.184 kJ. Therefore, the work done (in kJ) is:

W = 2.50 × 10^2 kcal × (4.184 kJ/kcal) = 1.046 × 10^3 kJ

Now, we can calculate the change in internal energy:

ΔU = Q - W
ΔU = -5.00 × 10^2 kJ - 1.046 × 10^3 kJ
ΔU = -1.546 × 10^3 kJ

The change in internal energy of the system is -1.546 × 10^3 kJ.

To find the change in internal energy of the system, we need to consider the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).

Given:
Q = -5.00 × 102 kJ (since the heat is released into the surroundings, it is negative)
W = 2.50 × 102 kcal

First, let's convert the units:
1 kcal = 4.184 kJ
2.50 × 102 kcal = 2.50 × 102 × 4.184 kJ = 1.046 × 104 kJ

Now we can calculate the change in internal energy:
ΔU = Q - W
= (-5.00 × 102 kJ) - (1.046 × 104 kJ)
= -5.00 × 102 kJ - 1.046 × 104 kJ
= -1.546 × 104 kJ

Therefore, the change in internal energy of the system is -1.546 × 104 kJ.

To find the change in internal energy of the system, we need to take into account the work done on the system and the heat released from the system. The change in internal energy (ΔU) can be calculated using 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.

Mathematically, the equation is:

ΔU = Q - W

Where:
ΔU = Change in internal energy
Q = Heat added to the system
W = Work done by the system

In this scenario, we are given that the work done on the system (W) is 2.50 × 10^2 kcal, and the heat released from the system (Q) is 5.00 × 10^2 kJ. However, these values need to be converted to the same unit before calculating the change in internal energy.

1 kcal = 4.184 kJ (conversion factor)

Converting the work done on the system from kcal to kJ:

W = 2.50 × 10^2 kcal × 4.184 kJ/kcal
W = 1050 kJ

Now that we have the values in the same unit, we can calculate the change in internal energy:

ΔU = Q - W
ΔU = 5.00 × 10^2 kJ - 1050 kJ
ΔU = -550 kJ

Therefore, the change in internal energy of the system is -550 kJ. The negative sign indicates that the system has lost energy.