during a process, 21.6 MJ of heat is added to closed system. if the internal energy is increased by 59.6 MJ, how much work in Btu was done? is the work done on ro by the system?
during a process, 21.6 MJ of heat is added to closed system. if the internal energy is increased by 59.6 MJ, how much work in Btu was done? is the work done on ro by the system?
To find the work done, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
Here, ΔU represents the change in internal energy, Q represents the heat added, and W represents the work done.
Given that the heat added to the system is 21.6 MJ and the change in internal energy is 59.6 MJ, we can use the equation above to find the work done.
ΔU = 59.6 MJ
Q = 21.6 MJ
Plugging in these values, we have:
59.6 MJ = 21.6 MJ - W
Rearranging the equation, we can solve for W:
W = 21.6 MJ - 59.6 MJ
W = -38 MJ
The negative sign indicates that work is done on the system, rather than by the system. The work done is 38 MJ.
To convert from MJ to BTU, we can use the conversion factor:
1 MJ = 947.8171 BTU
Therefore, the work done in BTU is:
38 MJ * 947.8171 BTU/MJ = 36,014.02 BTU
So the work done is approximately 36,014.02 BTU, and it is done on the system.