Do SI Units Eliminate the Need for Mechanical Equivalent of Heat?

[Nikhil Rajagopalan]

When computing the rise in temperature on a body due to mechanical work, if we stick to using SI units, do we need the conversion factor called mechanical equivalent of heat. That is, can we readily equate W = Q and hence W = m x C x ΔT . Where 'm' is the mass of the substance on which work is done, 'C' is its specific heat capacity and ' ΔT ' its the rise in temperature, all in SI units.

 

[Dale]

if we stick to using SI units, do we need the conversion factor called mechanical equivalent of heat.

The Joule is the SI unit of both work and heat, so there is no conversion factor in SI.

That is, can we readily equate W = Q

They have the same SI units, but they are still not the same thing in classical thermodynamics.

 

[tech99]

The Joule is the SI unit of both work and heat, so there is no conversion factor in SI.

They have the same SI units, but they are still not the same thing in classical thermodynamics.

The following light hearted video clip seems apposite :-

 

[sophiecentaur]

the conversion factor called mechanical equivalent of heat

It's present, in effect because the heating effect is determined by the heat capacity of the object in question. This is now stated in terms of Joules needed to raise its temperature, rather than Calories required to do the same thing.
I actually highly approve of the old term "Mechanical Equivalent of Heat" because it makes you remember that they have moved on from the Caloric Theory and actually spotted that Energy can have more than one form. Students who grow up on SI could miss out on the significance of the stunning discovery by Rumford, Joule and others.

 

[Dale]

The following light hearted video clip seems apposite :-

Very cute video! Cute, but wrong

 

[Mister T]

When computing the rise in temperature on a body due to mechanical work, if we stick to using SI units, do we need the conversion factor called mechanical equivalent of heat.

One joule of work is equivalent to one joule of heat. The equivalence is not just a unit conversion. It's a statement that two things that were previously thought to be different are instead the same.

That is, can we readily equate W = Q and hence W = m x C x ΔT . Where 'm' is the mass of the substance on which work is done, 'C' is its specific heat capacity and ' ΔT ' its the rise in temperature, all in SI units.

As long as you understand that those are two different processes. That is, you can have one process where you do work $W$ on an object and increase its internal energy by $cm\Delta T$. In which case you might write $\Delta U = W$. Or a different process where you transfer heat $Q$ to an object, increase its internal energy by $cm\Delta T$, and write $\Delta U = Q$.

$W$ in the first process is equal to $Q$ in the second process. The fact that they have the same effect on the object is the true meaning of the equivalence of heat and work.

The full generalization is the First Law of Thermodynamics, $\Delta U = Q+W$. (I'm using the convention where $W$ is the work done on a system.)