A thermal source supplies heat at a rate of 300 Js ⁻¹ . By using that heat a system performs work at a rate of 125 Js ⁻¹ . What is the rate at which the internal energy of the system increases?

Step-by-step solution

According to the first law of thermodynamics, the rate of heat supplied to the system is equal to the sum of the rate of work done by the system and the rate of change of internal energy of the system. Let dQ dt be the rate of heat supply, dW dt be the rate of work done, and dU dt be the rate of change of internal energy. The equation is given by dQ dt = dW dt + dU dt . Given values are dQ dt = 300 Js ⁻¹ and dW dt = 125 Js ⁻¹ . Substituting these values into the equation: 300 = 125 + dU dt dU dt = 300 - 125 = 175 J