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

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?

Change in internal energy of the system is 1800.5 J

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

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