The First Law of Thermodynamics — Question 8

Question 8

Time-dependent heat and work powers determine an energy maximum

Use $\Delta U=Q-W$, where $Q$ is heat into the system and $W$ is work done by it. For $0\le t\le10$ s, a closed stationary system receives heat at rate $\dot Q=100$ W and delivers work at rate $\dot W=20t$ W, where $t$ is in seconds. Initial internal energy is 2000 J.

1. Derive $U(t)$ and find its maximum value and the time at which it occurs.

2. Find the endpoint energy and cumulative heat and work. Does returning to the initial internal energy establish that every state variable has returned?

Question 8 - Solution

Time-dependent heat and work powers determine an energy maximum

(a) Method and calculation

$$\frac{dU}{dt}=100-20t\ \Rightarrow\ U(t)=2000+100t-10t^2.$$$$dU/dt=0\ \Rightarrow\ t_{\max}=5.00\,\mathrm s,\quad U_{\max}=2250\,\mathrm J.$$

(b) Interpretation and further calculation

$$Q_{0\to10}=100(10)=1000\,\mathrm J,\quad W_{0\to10}=\int_0^{10}20t\,dt=1000\,\mathrm J.$$$$U(10)=2000\,\mathrm J.$$The endpoint equality of one state function does not by itself establish a complete thermodynamic cycle; other independent state variables must also match.

Physical check

Internal energy starts falling when output power exceeds input power, even though heat continues to enter.