If 18oo cal:of heat is added to a system while system does work equivalent to 2800 cal:by expanding against its surroundings. What is the value of deltaE for the system?
So
dE=1800-2800
dE=-1000
Is this answer correct?
dE = q + w
That's what I would do.
To find the value of ΔE (change in energy) for the system, we can use the first law of thermodynamics:
ΔE = Q - W
Where ΔE is the change in energy of the system, Q is the heat added to the system, and W is the work done by the system.
In this case, the heat added to the system is 1800 cal and the work done by the system is 2800 cal.
To calculate the value of ΔE, we can substitute these values into the equation:
ΔE = 1800 cal - 2800 cal
Subtracting these values gives:
ΔE = -1000 cal
Therefore, the value of ΔE for the system in this situation is -1000 cal.