The heat supplied ( Q ) to a thermodynamic system and the work done ( W ) by the system are plotted against time ( t ). Both plots are straight lines passing through the origin. If the Q versus t line passes through the coordinates ( 4 s , 240 J ) and the W versus t line passes through ( 4 s , 80 J ), what is the rate of change of internal energy of the system?

Step-by-step solution

The rate of heat supplied is the slope of the Q versus t graph. Since it is a straight line passing through the origin (0, 0) and the point ( 4 s , 240 J ): dQ dt = 240 J - 0 4 s - 0 = 60 J/s = 60 W Similarly, the rate of work done is the slope of the W versus t graph, passing through the origin and ( 4 s , 80 J ): dW dt = 80 J - 0 4 s - 0 = 20 J/s = 20 W According to the first law of thermodynamics in rate form: dQ dt = dU dt + dW dt Substituting the calculated rates: 60 = dU dt + 20 dU dt = 60 - 20 = 40 W The rat