Question

If a system absorbs 300.5J of heat and performs 1.50kJ of work on the surroundings, what is the ΔE?

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Answer 1

The intern energy of the system is -1199.5 J

Step-by-step explanation:

Given:

System absorbs 300.5 J of heat

Performs 1.5 kJ = 1500 J of work on the surroundings

Question: What is the ΔE?

When a system absorbs heat, it will have a positive sign, therefore +300.5 J

When the system performs work on the surroundings, its sign will be negative, therefore, -1500 J

According the first law of thermodinamic:

ΔE = q + W = 300.5 - 1500 = -1199.5 J

Answer 2

Complete question:

If a system absorbs 300.5J of heat and performs 1.50kJ of work on the surroundings, what is the change in internal energy of the system ΔE?

Answer:

Change in internal energy of the system is 1800.5 J

Step-by-step explanation:

Given:

heat transferred to the system, Q = 300.5J

work done on the surroundings, W = 1.50kJ

Change in internal energy of the system ΔE, can be calculated by applying thermodynamic equation of change in internal energy, work done and heat transferred.

ΔE = Q + W

Where;

ΔE is change in internal energy

Q is heat transfer

W is work done

ΔE = 300.5 J + 1500 J

ΔE = 1800.5 J

Therefore, change in internal energy of the system is 1800.5 J