During a thermodynamic process, external work is performed on a system at a constant rate of 300 W . Simultaneously, the internal energy of the system is observed to decrease at a rate of 100 W . The rate of heat transfer and its direction are:

Step-by-step solution

According to the first law of thermodynamics, the rate of heat transfer is given by: dQ dt = dU dt + dW dt Using the standard sign convention: Work is done on the system, so dW dt = -300 W . Internal energy decreases, so dU dt = -100 W . Substituting these values: dQ dt = -100 W + (-300 W ) = -400 W The negative sign indicates that heat is being released by the system to the surroundings at a rate of 400 W . Answer: 400 W , released by the system