Consider the following experiment to determine the mechanical equivalent of heat $\mu = \frac{\Delta W}{\Delta Q}$
Not seen in the picture is a weight with a known mass hanging down the cord. When you start cranking, you compensate the weight force $F_G$ with the friction $F_F$. Its value is the difference of $F_G$ and the force held by the dynamometer.
One can calculate the work put into the system via $\Delta W = F_F \cdot\Delta s = (F_G - F_D) \cdot 2\pi rn$, with $r$ the cylinder radius and $n$ the number of revolutions. (you don't have to consider vectors calculating $\Delta W$, because $F_F$ and $s$ always point to the same direction $\Rightarrow$ $\phi = 0 \Rightarrow \cos(\phi) = 1$)
By measuring the temperature, one can obtain $\Delta Q = C_{total}\cdot\Delta T$.
To the questions. What are the possible outcomes for $\mu$? In an idealized case, where no energy gets lost/dissipated, one can clearly assume $\mu = 1$. What's the meaning behind $\mu > 1$ and $\mu < 1$?
My ideas and assertions:
- In a closed, ideal system, one can completely transfer all mechanical energy $W$ in heat energy $Q$
- One cannot transfer all heat $Q$ back into $W$ (Perpetuum mobile 2nd kind)
This yields, $\Delta Q$ must always be greater than $\Delta W$, because you can't transfer everything into mechanical work $\Rightarrow \mu \leq 1$. But, viewed the other way round, if $\Delta W$ is less than $\Delta Q$, where does the energy to raise $Q$ come from? The cylinder is in thermal equilibrium with the room temperature before one starts to crank, so there is no input from another warmth bath.