A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal energy in tank.

U₂ = 400 KJ

Explanation:

Given that

Initial energy of the tank, U₁ = 800 KJ

Heat loses by fluid, Q = - 500 KJ

Work done on the fluid, W = - 100 KJ

Sign -

1. Heat rejected by system - negative

2. Heat gain by system - Positive

3. Work done by system = Positive

4. Work done on the system-Negative

Lets take final internal energy = U₂

We know that

Q = U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = - 500 + 900 KJ

U₂ = 400 KJ

Therefore the final internal energy = 400 KJ