First Law of Thermodynamics Problem 1. The heat transfer loss in a non-flow process constitute 1200 kJ meanwhile an internal energy increases by 250 kJ. The problem is to obtain the work transfer and state of the process (expansion or compression). Solution: A system which is being considered is closed. Therefore the first thermodynamics low could be applied, which is as follows: Q− W = Δ U , (1) where Q is a heat transfer loss; W is the work transfer; ΔU is difference in internal energy. From the equation (1) above the work transfer is: W = Q− Δ U = − 1200− 250= − 1450 kJ
The work transfer is: W=-1450kJ The negative value of the work means that the state of process is compression. Problem 2. A specific internal energy decrease of 70 kJ per kg was observed in a non-flow process which was carried out on 6,2 kg of a substance. A work transfer from the substance is 60 kJ per kg. The problem is to find out if the heat transfer is gain or loss by determining its value. Solution: We consider that the system in the problem is closed. Therefore the First low of Thermodynamics could be applied: Q− W = Δ U , (1) From the equation (1) the value of the heat transfer could be found:
Q= W + Δ U = 6,2∗ (− 70)+ 6,2∗ 60= − 62kJ , Therefore the heat transfer is Q=-62kJ. The value is negative, therefore the conclusion could be made that the heat is exiting the control volume and it is loss. Problem 3. 200 kJ/kg of the heat from the working substance was transferred out of the system during the working stroke of an engine. The engine makes work of 300 kJ/kg. The problem is to determine the change in internal energy and identify if the state is decrease or increase. Solution: