If a system has 4.50×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (ΔE or Δ) of the system?

The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.

Given the following data:

Conversion:

1 kcal = 4.184 kJ

=

To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.

Mathematically, the first law of thermodynamics is given by the formula:

Where;

Substituting the given parameters into the formula, we have;

Change in internal energy, E = -18823 Kilojoules

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Explanation:

The internal energy of the system according to the thermodynamics law is the difference between the heat released by the system and work done to the system. Mathematically;

ΔE = q - W where;

ΔE is the change in internal energy of the syatem

q is the heat released by the system

W is the work done to the system

Given W = 4.50×10² kcal and q = 5.00×10² kJ

Since 1cal = 4.184Joules

4.50×10² kcal = (4.50×10²×10³)×4.184

W = 4.50×10² kcal = 1,882,800 Joules

q = 5.00×10² kJ =  5.00×10⁵Joules

q = 500,000Joules

Since ΔE = q - W

ΔE = 500,000 J -  1,882,800 J

ΔE  = -1,382,800Joules

Hence, the change in the internal energy of the system is -1,382,800Joules

c, 1a hope this

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000