The amount of heat flow out of the system if a work of 2550 j is done on an ideal gas, but the internal energy increases by only 1500 j is - 1050 J
ΔU = Q + W
ΔU = Change in energy
Q = Heat
W = Work done
W = 2550 J
ΔU = + 1500 J
Positive because it is mentioned that internal energy increases.
Q = ΔU - W
Q = 1500 - 2550
Q = - 1050 J
Here the amount of heat flow is 1050 J. Since it is negative heat flows out of the system. If the value of Q is positive, then the heat flow is into the system.
Therefore, the amount of heat flow is 1050 J out of the system.
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