# Mechanical equivalent of heat => transfer from work $W$ to heat $Q$

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.

First, I should say that $$\Delta W$$ and $$\Delta Q$$ don't have meaning. $$\Delta$$ exists for state functions, but work and heat are path functions. A system doesn't have heat or work. Heat and work are recognized when they are transmitted through system's boundary.

Second, the correct equation is $$\Delta U=mC\Delta T$$. So, I assume that you want to determine $$\large{\frac W{\Delta U}}$$

What's the meaning behind $$\mu\gt1$$ and $$\mu\lt1$$?

$$\mu\gt1$$ means some portion of work has been used for another purpose (for example, increasing surrounding temperature) except increasing the internal energy of your sightly system.

$$\mu\lt1$$ means the system has received some heat from surrounding.

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)

I cannot understand what is the $$Q$$. As I mentioned a system doesn't have heat. During the first experiment, work is converted to internal energy of the system. Now, there is no heat here. System has a internal energy bigger than its initial state.

• Note that the letter "$C$" is used for both heat capacity and specific heat capacity, so the OPs equation can be correct. Sometime authors use uppercase "$C$" for heat capacity and lower case "$c$" for specific heat, but you can't depend on that. Nov 1, 2016 at 14:10