If a system has 375 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 ΔU) of the system?

The change in internal energy (ΔE or ΔU) of the system is 1069 kJ.

Given data:

The amount of work done is, W = - 375 kcal.  (Negative sign shows that work is done on the system)

1 kcal = 4.184 kJ

Then,

The Amount of heat released is,  (Negative sign shows that heat is released)

Use first law of thermodynamics which says that the amount of heat given to the system is equal to the work done and internal energy change. That is,

Q = W + U

Here, U is the internal energy  change.

Solving as,

Thus, the change in internal energy (ΔE or ΔU) of the system is 1069 kJ.

Learn more about the first law of thermodynamics here:

link

  ΔU= 1069 kJ

Explanation:

Given that

The work done on the system

W= 375 kcal                      ( 1 kcal = 4.184 kJ)

W= - 1569 kJ

Heat release by system ,Q = 5 x 10² kJ

Q= - 500 kJ

Sign convention :

Work done on the system is taken as negative and work done by the system is taken as positive .

Heat gain by system is taken as positive and heat release by system is taken as negative.

Now by using first law of thermodynamics

Q= W+ ΔU

Q=Heat transfer

W=Work

ΔU =Change in the internal energy

Now by putting the values

-500 = -1569 + ΔU

ΔU= -500+ 1569 kJ

ΔU= 1069 kJ

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