In a certain chemical process,a lab technician supplies 254 j of heat to a system. At the same time,73 j of work are done on the system by iys surroundings.What is the increase in the internal energy of the system?
254-(-73)=327
Since heat is supplied to a system Q=+ve = + 254J = 254J
Since workdone on the system so workdone is negative
Then W= - 73 J
Then change in internal energy ∆U= 254-(-73) = 327 Joule
To find the increase in the internal energy of the system, we need to consider the heat supplied and the work done.
The change in internal energy (ΔU) can be calculated using the first law of thermodynamics:
ΔU = Q - W
Where:
ΔU = Change in internal energy
Q = Heat supplied to the system
W = Work done on the system
Substituting the given values, we have:
ΔU = 254 J - 73 J
ΔU = 181 J
Therefore, the increase in the internal energy of the system is 181 J.
To determine the increase in the internal energy of the system, you need to consider the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat supplied to the system (Q) minus the work done by the system (W):
ΔU = Q - W
In this case, the heat supplied to the system is 254 J (Q = 254 J) and the work done on the system is 73 J (W = 73 J). Now, we can substitute these values into the equation to find the change in internal energy:
ΔU = 254 J - 73 J
ΔU = 181 J
Therefore, the increase in the internal energy of the system is 181 J.