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FIRST LAW OF THERMODYNAMICS FOR NONFLOW PROCESS
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Question 1 of 4
1. Question
1 pointsA system consisting of some fluid is stirred in a tank. The rate of work done on the system by the stirrer is 2.25 hp. The heat generated due to stirring is dissipated to the surroundings. If the heat transferred to the surroundings is 3400 kJ/h, determine the change in internal energy(answer upto three decimal)
Correct
Incorrect

Question 2 of 4
2. Question
1 pointsA system consisting of a gas confined in a cylinder is undergoing the following series of processes before it is brought back to the initial conditions:
Step 1: A constant pressure process when it receives 50 J of work and gives up 25 J of heat.
Step 2: A constant volume process when it receives 75 J of heat.
Step 3: An adiabatic process. (Note: In adiabatic process there is no heat exchange between system
and surroundings.)
Determine the change in internal energy during each step and the work done during the adiabatic process.
Correct
The process is shown in Fig. below AB is a constant pressure process
Internal energy change for this step is calculated as
(DU)AB = QAB – WAB = – 25 + 50 = 25 J
During step BC, no work is done, as it is a constant volume process. Change in internal energy is
(DU)BC = QBC = 75 J
Step CA being an adiabatic process, QCA = 0. Therefore,
(DU)CA = – WCA
For a cyclic process, the net change in internal energy is zero. That is,(DU)AB + (DU)BC + (DU)CA = 0
(DU)CA = – [(DU)AB + (DU)BC] = – (25 + 75) = – 100 J
Therefore, WCA, the work done during the adiabatic process = – (DU)CA = 100 J.Incorrect
The process is shown in Fig. below AB is a constant pressure process
Internal energy change for this step is calculated as
(DU)AB = QAB – WAB = – 25 + 50 = 25 J
During step BC, no work is done, as it is a constant volume process. Change in internal energy is
(DU)BC = QBC = 75 J
Step CA being an adiabatic process, QCA = 0. Therefore,
(DU)CA = – WCA
For a cyclic process, the net change in internal energy is zero. That is,(DU)AB + (DU)BC + (DU)CA = 0
(DU)CA = – [(DU)AB + (DU)BC] = – (25 + 75) = – 100 J
Therefore, WCA, the work done during the adiabatic process = – (DU)CA = 100 J. 
Question 3 of 4
3. Question
1 pointsA gas is undergoing a change of state from A to B along path ACB in which the total heat supplied to the system is 80 J and the work done by the system is 30 J (Fig. below). The cycle is completed by bringing the system back to the initial state along the curved path BA for which work of 40 J is done on the system.
The heat quantity involved in the process AD and DB, if the internal energy of the system at state D is greater than that at state A by 40 J.
Correct
For the process occurring along ACB, Q = 80 J, W = 30 J. Q is positive since heat is supplied to the system and W is positive since work is done by the system. The first law of thermodynamics gives
DU = Q – W
Therefore, the change in internal energy for the process ACB is
DU = 80 – 30 = 50 JNo work is involved in process AD since it is a constant volume process. The change in
internal energy in process AD is given to be DU = 40 J. Therefore, heat involved in this process
is
Q = DU + W = 40 + 0 = 40 J (Heat is supplied to the system.)
Change in internal energy between states B and A is calculated to be 50 J. Now,
DUAB = DUAD + DUDB
or
DUDA = DUAB – UAB = 50 – 40 = 10 J
Work done by the system in process DA is 3 times the work done along path AC =
90 J. Therefore, heat quantity involved is
Q = DU + W = 10 + 90 = 100 J (Heat is supplied to the system.)Incorrect
For the process occurring along ACB, Q = 80 J, W = 30 J. Q is positive since heat is supplied to the system and W is positive since work is done by the system. The first law of thermodynamics gives
DU = Q – W
Therefore, the change in internal energy for the process ACB is
DU = 80 – 30 = 50 JNo work is involved in process AD since it is a constant volume process. The change in
internal energy in process AD is given to be DU = 40 J. Therefore, heat involved in this process
is
Q = DU + W = 40 + 0 = 40 J (Heat is supplied to the system.)
Change in internal energy between states B and A is calculated to be 50 J. Now,
DUAB = DUAD + DUDB
or
DUDA = DUAB – UAB = 50 – 40 = 10 J
Work done by the system in process DA is 3 times the work done along path AC =
90 J. Therefore, heat quantity involved is
Q = DU + W = 10 + 90 = 100 J (Heat is supplied to the system.) 
Question 4 of 4
4. Question
1 pointsA gas is undergoing a change of state from A to B along path ACB in which the total heat supplied to the system is 80 J and the work done by the system is 30 J (Fig. below). The cycle is completed by bringing the system back to the initial state along the curved path BA for which work of 40 J is done on the system.
The heat supplied or removed in the process along path BA.
Correct
For the process occurring along ACB, Q = 80 J, W = 30 J. Q is positive since heat is
supplied to the system and W is positive since work is done by the system.
The first law of thermodynamics gives
DU = Q – W
Therefore, the change in internal energy for the process ACB is
DU = 80 – 30 = 50 JFor the return path along BA, W = – 40 J, DU = – 50 J (negative of the change in internal energy
for the process from A to B.)
Q = DU + W = –50 – 40 = –90 J (Heat is given out by the system.)Incorrect
For the process occurring along ACB, Q = 80 J, W = 30 J. Q is positive since heat is
supplied to the system and W is positive since work is done by the system.
The first law of thermodynamics gives
DU = Q – W
Therefore, the change in internal energy for the process ACB is
DU = 80 – 30 = 50 JFor the return path along BA, W = – 40 J, DU = – 50 J (negative of the change in internal energy
for the process from A to B.)
Q = DU + W = –50 – 40 = –90 J (Heat is given out by the system.)