Test: First Law of Thermodynamics - 2 - Mechanical Engineering MCQ

First law of thermodynamics is valid for

During a thermodynamic process, 84 kJ of heat flows into the system and the work done by the system is 32 kJ. The increase in internal energy of the system is

The specific heat at constant pressure for an ideal gas is given by 
c_p = 0.9 + (2.7 x 10^-4) T (kJ/kgK)
Where T is in kelvin. The change in enthalpy for this ideal gas undergoing a process in which the temperature changes from 27°C to 127°C is most nearly.

A closed system undergoes a process 1-2 for which the values of Q_1-2 and W_1-2 are +20 kJ and +50 kJ respectively. If the system is returned to state 1 and Q_2-1 is - 10 kJ what is the value of work W_2-1

Given that along the path 1-2-3 a system absorbs 100 kJ as heat and does 60 kJ work while along the path 1 -4-3 it does 20 kJ work (see figure given). The heat absorbed during the cycle 1-4-3 is

The network output for the cycle 1-2-3-4 -5-6-1 shown in figure is

In a reversible isothermal expansion process fluid expands from 10 bar and 2 m³ to 2 bar and 10 m³. During the process the heat supplied is at the rate of 100 kW. What is the rate of work done during the process

The state of an ideal gas is changed from (T₁, P₁) to (T₂, P₂) in a constant volume process. To calculate the change in enthalpy (Δh) ail of the following properties/variables are required.

For the two paths as shown in the figure, one reversible and one irreversible, to change the state of the system from a to b.

The net work done for the closed system shown in the given pressure-volume diagram is

Two ideal heat engine cycles are represented in the given figure. Assume VQ = QR, PQ = QS and UP = PR= RT. If the work interaction for the rectangular cycle (WVRU) is 48 Nm, then the work interaction for the other cycle PST is

A mixture of gases expands from 0.03 m³ to 0.06 m³ at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is