What is the efficiency of the energy conversion in the cannon bore?

[Fantini]

The problem:

Rumford observed that when one horse turned a cannon bore for about 2.5 hr the temperature of 27 lb of water that was initially ice cold (i.e., near 0°C) rose to near boiling hot (i.e., near 100°C). Given that one horsepower is 33,000 ft-lb;min, that 1 kg weighs 2.2 lb, and that 1 in. = 2.54 cm, what is the mechanical equivalent of heat in joules per calorie implied by these rough numbers?

I converted all of the units to SI and equated $Q = mc \Delta T$, trusting that we have to find is $c$. The numeric value I found for $c$ is 5.468 cal/g°C. The answer is 5.48 J/cal.

The numbers match, but I'm not sure I'm understanding the concept behind what I'm doing. The mechanical equivalent of heat is how much work you have to perform on the system (or by the system) to generate the same temperature difference due to heat. How would I compute that in this case?

 

[I like Serena]

The problem:

Rumford observed that when one horse turned a cannon bore for about 2.5 hr the temperature of 27 lb of water that was initially ice cold (i.e., near 0°C) rose to near boiling hot (i.e., near 100°C). Given that one horsepower is 33,000 ft-lb;min, that 1 kg weighs 2.2 lb, and that 1 in. = 2.54 cm, what is the mechanical equivalent of heat in joules per calorie implied by these rough numbers?

I converted all of the units to SI and equated $Q = mc \Delta T$, trusting that we have to find is $c$. The numeric value I found for $c$ is 5.468 cal/g°C. The answer is 5.48 J/cal.

The numbers match, but I'm not sure I'm understanding the concept behind what I'm doing. The mechanical equivalent of heat is how much work you have to perform on the system (or by the system) to generate the same temperature difference due to heat. How would I compute that in this case?

The units are different, so if the numbers are close, that is merely coincidence.
It means we're talking about different quantities.

Btw, the unit of your horsepower should be $\text{ft}\cdot\text{lbf}/\text{min}$, where $1\text{ lbf} = 1 \text{ lb} \cdot g =1 \text{ lb} \cdot 9.81 \frac{\text{m}}{\text{s}^2}$.

The formula to calculate the work of the horse is:
$$W=P\cdot t$$
where W is work, P is power, and t is time.

From $W=P\cdot t$ and the proper conversions, we get the work done by the horse in joules.
From $Q=mc\Delta T$, we get the same energy in calories.

Btw, the value of $c$ was apparently once defined as $1 \text{ cal}/(\text{g}\cdot\text{°C})$.
Where did you get $5.468 \text{ cal}/(\text{g}\cdot\text{°C})$ from?

Set them equal to each other, and we can calculate the ratio between joules and calories, which should come out as a constant that depends on which definition for calorie we use. In all cases it is:
$$4.182 \lessapprox \frac{\text{cal}}{\text{J}} \lessapprox 4.204$$

 

[Fantini]

What I did was convert $27$ lb to $kg$, finding $m = 27/2.2 \text{ kg} = 27/2.2 \cdot 10^3 \text{ g}$. I used the same formula you did for energy, finding $$W = \frac{33000 \text{ ft-lb}/\text{min}}{0.7376 \text{ ft-lb}/\text{J}} \cdot 150 \text{ min} = \frac{33000 \cdot 150}{0.7376} \text{ J}.$$ Then $Q=mc \Delta T$ implies $$c = \frac{Q}{m \Delta T} = \frac{33000 \cdot 150}{0.7376} \cdot \frac{1}{\frac{27}{2.2} \cdot 10^3} \cdot \frac{1}{100} = 5.468 \frac{\text{ J}}{\text{g} \cdot {}^{\circ} \text{C}}.$$

Edit: From the computations above I see that it's $5.468$ Joules per gram centimeters.

Does this mean that to arrive that the desired answer we divide by $$c_{\text{water}} = 1 \frac{\text{ cal}}{\text{g} \cdot {}^{\circ} \text{C}}$$ to arrive at the mechanical equivalent?

 

[I like Serena]

What I did was convert $27$ lb to $kg$, finding $m = 27/2.2 \text{ kg} = 27/2.2 \cdot 10^3 \text{ g}$. I used the same formula you did for energy, finding $$W = \frac{33000 \text{ ft-lb}/\text{min}}{0.7376 \text{ ft-lb}/\text{J}} \cdot 150 \text{ min} = \frac{33000 \cdot 150}{0.7376} \text{ J}.$$

That should be $\text{lbf}$ instead of $\text{lb}$. (Worried)
The difference is a factor of $9.81$ and a different unit.
$1 \text{ lbf}$ is the weight of $1 \text{ lb}$, which is the force with which the Earth pulls on the mass. (Nerd)You're only getting away with it, because the units cancel. (Whew)

Does this mean that to arrive that the desired answer we divide by $$c_{\text{water}} = 1 \frac{\text{ cal}}{\text{g} \cdot {}^{\circ} \text{C}}$$ to arrive at the mechanical equivalent?

Yes. (Nod)

More accurately, you're using the mechanical equivalent to approximate the ratio between joule and calorie.
We can expect some energy to "leak" away, which is why your value is higher than the defined value. (Nerd)

 

[Fantini]

That should be $\text{lbf}$ instead of $\text{lb}$. (Worried)
The difference is a factor of $9.81$ and a different unit.
$1 \text{ lbf}$ is the weight of $1 \text{ lb}$, which is the force with which the Earth pulls on the mass. (Nerd)You're only getting away with it, because the units cancel. (Whew)

Then the book is wrong, as I copied verbatim. (Worried) But I understand your point now.

More accurately, you're using the mechanical equivalent to approximate the ratio between joule and calorie.
We can expect some energy to "leak" away, which is why your value is higher than the defined value. (Nerd)

You mean that we measure additional energy that "leaked" from somewhere else, hence why the value is higher?

 

[I like Serena]

You mean that we measure additional energy that "leaked" from somewhere else, hence why the value is higher?

The power the horse provides makes the temperature of the water rise, but it also makes the area around the water rise, which leaks away without use to us.

More specifically, the efficiency of the energy conversion in the cannon bore is:
$$\text{efficiency} = \frac{\text{Useful output energy}}{\text{Total input energy}}
=\frac{Q}{W} = \frac{4.184}{5.468} = 76.5 \%$$