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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?

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

Given the following data:

  • Quantity of heat = [tex]5.00 \times 10^2 \;kJ[/tex]
  • Work done = [tex]4.50 \times 10^2 \;kcal[/tex]

Conversion:

1 kcal = 4.184 kJ

[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]

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:

[tex]\Delta E = Q - W[/tex]

Where;

  • [tex]\Delta E[/tex] is the change in internal energy.
  • Q is the quantity of heat released.
  • W is the work done.

Substituting the given parameters into the formula, we have;

[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]

Change in internal energy, E = -18823 Kilojoules

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