Thermal Equilibrium

After seeing this picture, my friend texted me asking if we could try this, so we set up our own using hot coffee, a glass of ice water, and straws.
The principle it's based upon is thermal equilibrium, but I didn't think the ice water would be enough to actually cool the coffee a significant amount since it moved so quickly through the straw—It did actually work.

Theoretically, the calculations for the thermal equilibrium equation would be as follows:

Assuming there is approximately one straw full of coffee in the ice water, and the average drinking straw has a length of around 8 inches and a diameter of 0.21 inches, the volume of the coffee inside the straw will be 4.541x10(^-6) m³. Since coffee is very similar to water, we can use the conversion 1 m³ of water = 1000.0 kg of water, giving us a mass of .0045 kg of coffee in the straw. Since we got the coffee fresh, and most coffee is served at about 180°F, we convert that to 82.2°C. 

The straw only takes up 4.541x10(^-6) m³ in the water, so it won't interact with the entirety of the cup of water. Using double the volume of water displaced by the straw, to represent the volume surrounding the straw that might have a chance of interacting with the coffee in the straw in such a short amount of time, gives us .009 kg in the cup at freezing. 

So Qʷᵃᵗᵉʳ+Qᶜᵒᶠᶠᵉᵉ = 0J
Because we use water for both of them, specific heat (c) can be cancelled out.

(.0045kg)(cwater)(Tᶠ - 82.2) + (.009kg)(cwater)(Tᶠ - 0) = 0J
.0045Tᶠ - .3699 + .009Tᶠ = 0J
.0135Tᶠ = .3699
Tᶠ = 27.4°C

Ice cream heat transfer

The other day, I set out a gallon of ice cream so it could thaw enough for me to actually be able to scoop some out with a plastic fork, because I am a lazy college student who doesn't want to do the dishes. In considering what this melting entails, I thought about heat transfer.

In order for the ice cream to melt, heat is is absorbed. Enough heat needs to be absorbed to first melt the ice cream, Lf for ice cream, and then heat continues to absorb so that the ice cream can reach thermal equilibrium with its surroundings. Obviously, no one wants to eat room temperature ice cream. Therefore, the ideal situation is to scoop the ice cream before it can absorb enough heat to completely melt. The energy being transferred to melt the ice cream comes from the surroundings such as the counter or air. It also explains why people will run an ice cream scooper under hot water prior to scooping. The warm scooper provides energy necessary to melt the ice cream, and make it easier to scoop. Therefore, it's unfortunate I am a broke college student and don't own an ice cream scooper.

Unfortunately, I didn't have a scale to weigh the half empty ice cream container, but the heat of fusion for vanilla ice cream is about 204kj/kg, meaning it's okay if you forget about the ice cream for a few minutes, as it takes quite a bit of energy to melt. According to the internet, it takes more energy to melt vanilla ice cream than chocolate or strawberry. I don't really have an explanation for this, but just thought it was an interesting reasoning for next time your friend's chocolate ice cream is melting faster than your vanilla.

Curlgate

If you’ve been keeping up on your Physics 111 blog post readings, you’d know that Emily Schweitzer is the fearless leader of Colgate’s Curling team. I, although not nearly as skilled, am also a curler, and am fascinated by the physics that makes curling work (like nearly perfect elastic collisions, angular motion, etc.). If you need a basic introduction to the sport as a whole, as it’s important for my post, here’s a wonderful crash course: https://www.youtube.com/watch?v=IOk9SVzqHsk

There has recently been significant strife in the curling community about certain synthetic fabrics that have been used to make the broom heads for sweeping. These new broom heads have different properties such as less waterproofing, artificial textures/hairs, and different kinds of padding, all which have the ability to melt and carve into the pebbles on the ice much more effectively than the standard broom heads. The concern here is that this new technology allows for cheating and lessening the skill required to curl well, as rocks can be manipulated much more when the brooms are much more effective at changing the ice in its path.

The main logic as to why the sweeping helps the stone travel farther is that by momentarily raising the temperature of the ice, but not melting it, you reduce the kinetic coefficient of friction, slowing the stone down less. You do not want to melt the pebbles on the ice, because then you are increasing the amount of surface area the stone has in contact with the ice, increasing the frictional force on the stone. 

I wanted to more thoroughly understand why these different types of broom heads change the ice in different ways. The standard broom head is padded and made of nylon coated in a waterproof material, and in this example I will examine what would happen if you swapped the fabric out for something that creates more friction with less padding, as done in the newly popular icePad broom head (which is banned in some levels of competition). This is supposed to increase the overall power of the sweep by reducing the surface area wasted on the unpebbled ice, allowing all of the frictional force to be concentrated on the pebbles. This change in fabric is also shown to create more friction between the broom and the ice to heat up the ice more, causing less friction between the stone and the ice. These two manipulations allows a sweeper to exert far less energy to manipulate the stone much more effectively.

If you're looking to read more on this, here’s a few articles that discuss this issue more in depth:

https://www.nytimes.com/2015/11/20/sports/facing-control-issues-curling-draws-line-at-high-tech-brooms.html

http://www.cbc.ca/sports/olympics/winter/curling/curling-s-broom-controversy-can-t-be-swept-aside-1.3448870

As for now, the verdict on new broom technology is as follows: use a standard broom, and there’ll be good curling!

I was foam rolling the other day and wondered if it mattered whether or not you had both legs on the roller at once when rolling out you quads. Logically it makes sense that you are doubling the weight and doubling the area on which that force is acting. But this can also be thought of in terms of pressure, or force/surface area. Because putting a second leg under the foam roller while the force remains the same (as body weight and gravity are not changing) changes the surface area, the total amount of pressure felt is less.

Here is an example with rough estimates:

Surface area of one leg (the amount of thigh that would be of the roller) about 20 cm² or 0.002m²

Force of thigh on roller: The thigh is estimated to be 12% of body weight in females (10.5% in males). We will assume the person weighs 130 pounds. 130 x 0.12 = 15.6 pounds or 0.45 kg

P= F/A = (0.45kg)(9.8 m/s²)/0.002m²

 = 2205 Pa

Both legs:

            P = 2(.45kg)(9.8 m/s²)/(0.002m²)2

            = 2205 Pa

Through these calculations, you can see that the same amount of pressure is felt regardless of having one or both legs on the roller.

The Physics of Salting Roads

Winter is a harsh season here in Colgate and usually, it snows heavily and makes it hard to walk or drive on the road. The recent snow up to now has not caused large congestion on the road because there are almost no snow or ice on the road. It is because a certain substance is pulled onto the road and lower the freezing point of ice, and it turns out to be very safe in the winter! Usually people add saltwater and it turns out to be very useful!

According to National Snow & Ice Data Center, saltwater has a relatively lower freezing point than freshwater, which freezes at 0 degree Celsius. And for saltwater,  its freezing point decreases by 0.28 degree Celsius for every 5 ppt increase in salinity. 

The graph above shows the freezing point of salt with respect to its salinity. By adding certain portion of salt into fresh water, the freezing point continuously decreases till around -21 degree Celsius when the salinity is around 290g/kg. After this point the freezing point begins to increase as the salinity increases.

The Physics of Santa

With Christmas right around the corner, I’ve decided to take a closer look at the physics of Santa as he travels to households across the world. The first thing I looked at is how fast Santa would need to go in order to deliver a present to every christian household with at least one good child. This can be determined if we say that Santa must travel 510,000,000km on Christmas Eve in 32 hours. That means that Santa would be traveling 1.6 x 10^5km/hr. This velocity is obviously impossible since we don't even have any mode of transportation that can travel that fast. Thus, making it impossible for Santa to deliver all the presents.

There are other factors that should be taken into account when looking at Santa’s travels during Christmas Eve. For example, the weight of the sled would be enormous because of all the toys that would need to be on the sled and a large number of reindeers would be needed in order to carry this massive amount of toys. Furthermore, since we are traveling in the x-direction we need to take into account that there would be a force of air resistance acting on Santa, the sled, and all the reindeer. This would need to be taken into account to find the force needed to move in the x direction. When we think about these factors, it makes the possibility of Santa being real impossible and that is why I looked into the physics of Santa.  

Coffee accident

This morning I walked into the library cafe and got my daily morning coffee. As I sat down, I slipped and the coffee spilled onto the table. I left the coffee there for a while, and then cleaned it up after. I started to ask myself how much heat energy was transferred from the coffee to the table for the coffee to lose so much heat.

The Coffee was approximately 71.5 C when I got it, and then fell to a room temperature of  23 C.

The mass of the Coffee Spilled was .10kg, and the specific heat for coffee and milk is 4.184 J/ (g C˚).

Using the equation :

Q= mc∆T

We can find that the transfer of heat energy is...

Q= (.10kg)( 4.184 J/ (g C˚))(48.5 C)

Q=20.3 J of energy is lost from the coffee to it's environment.

Radical Rotations

As I looked out my window this morning and saw the snow I couldn't help but look forlornly at my longboard leaning up in a corner of my room. Winter is here and I won't be riding that for quite sometime. But then I started thinking about the fact that a longboard can get going pretty quick even though it has relatively small wheels. That got me thinking, if I'm cruising down the hill how fast are those little wheels spinning?

The speed limit up the hill is fifteen miles per hour and going down Oak drive one can easily hit that because of the slope. This is equal to roughly 6.7 m/s. If the radius of the longboard wheels is 3.25 cm then we can convert these quantities into rotational velocities.

v=15 mph=6.7 m/s

r=3.25 cm=0.0325 m

w=v/r

w=(6.7 m/s)/(0.0325 m)=206 rad/s

We can take this angular velocity and turn it into revolutions per minute by converting our units.

(206 rad/s)(1 rev/2*pi rad)(60 s/1 min)=1967 rev/min

As a comparison: the average car runs at around 2000 rpm when cruising easily down the road. 

In any event, bombing the hill is probably a terrible idea and if we've learned anything from professor Metzler it's that you should ALWAYS wear a helmet.

Skydiving and Wingsuits

The other day I saw a video of a person who was wingsuit flying and I wanted to know some of the physics behind it. First, we can look at how a skydiver who is dropping vertically. When he falls toward the earth, he accelerates and his velocity increases. However, he will eventually reach a point where he is no longer accelerating and reaches his terminal speed. This is due to the fact that at this point his drag force is equal to his force of gravity. Up until this point, his force of gravity was greater than his drag force causing him to accelerate. Since Fg = Fd, and Fd is equal to ½Cd⍴airAv2 where Cd is the drag coefficient ⍴air is the density of air A is the cross sectional area and v is velocity, we can solve for the terminal velocity.

½Cd⍴airAv2 = mg

v = √(2mg/(Cd½⍴airA))

When skydiving, as well as wingsuit flying, a person is able to manipulate their terminal speed by changing their cross sectional area. One can go faster by pointing their head straight down and their legs up so that their area is as small as it can be. If they lay perpendicular to the drag force, their area will be maximized and their terminal velocity will be slower. In the case of wingsuit flying however, the design of a wingsuit allows for the velocity in the direction of the force of gravity to be converted to horizontal velocity. Lift force also affects wingsuit flight which helps them fly horizontally. Glide ratio is determined based on the relationship between lift, drag, and weight, or force of gravity. Most wingsuits have a glide ratio of 2.5:1 which means that they are able to move horizontally 2.5 m for every 1 m dropped vertically. Overall, lift is the main force that affects the horizontal flight which normal skydivers cannot achieve.

References:

https://adventure.howstuffworks.com/wingsuit-flying1.htm