I wanted to make a quick dinner last night so that I could get back to studying, so I decided to make pasta. Wanting to multitask, I did some dishes while the pasta was cooking in the boiling water. When I was done with the dishes I stirred the pasta with the fork that I had left in the pot. I forgot that the fork would probably be hot since I had left it in the boiling water. 

By the time I touched the fork after leaving it in the boiling water, its temperature would have been:

Q = mcΔT

The heat required to warm up an iron pot to 100 °C from room temperature is:

Qpot = 1 kg * 450 J/kg°C * (100 °C - 20 °C) = 36000 J

The heat required to take the water from room temperature to boiling is:

Qboilingwater= 0.25 kg * 4186 J/kg°C * (100 °C - 20 °C) = 83720 J

Qpot + Qboiling water = 119,720 J

The temperature change in an aluminum spoon from room temperature assuming that this is a closed system:

Q = mcΔT

119,720 J = 0.3 kg * 900 J/kg°C * (Tf - 20 °C)

Tf = 443 °C

This would have been hot enough to burn my hand. Luckily it was not a closed system and a lot of heat was lost to the air and I was able to go back to making notes pain-free.

Winter Weather Insulation

I am currently in a ecology class where I learn about the exchanges of energy in the atmosphere. We learned about radiation and how that effects the earth and soil. We also learned that snow reflects a lot of the radiation thus keeps the grass below cool, but it also keeps the grass insulated because snow is mainly air and air is a good insulator. What I want to determine is whether the albedo effect of the snow offsets the insulative effect of air gaps inside snow. 

Snow forms when it is 0 degrees C.

Assume there is 0.5m of snow.

Assume also that the area of the measured plot is 0.2m by 0.3m.

Thermal conductivity of snow is 0.05- 0.7 WK-1m-1

Assume that soil temperature is 39.7 degrees F (4.28 degrees C) in December (when it begins to snow)

Thus, using the lower scale of thermal conductivity, 

Q/T = (k * A * (T2 - T1)) / l

Q/T = (0.05 * 0.2 * 0.3 * (4.28 - 0)) / 0.5

Q/T = 0.02568

Using the higher scale of thermal conductivity, 

Q/T = (k * A * (T2 - T1)) / l

Q/T = (0.7 * 0.2 * 0.3 * (4.28 - 0)) / 0.5

Q/T = 0.35952

Emissivity constant of snow is 0.969 - 0.997. 

Boltzman Constant is 5.67 X 10^-8

T1 = 273K

T2 = 312.7K

Now, looking at albedo effects, (using the lower end of emissivity constant)

Q/T = eoA(T1^4 - T2^4)

Q/T = (0.969)(5.67 X 10^-8)(0.2 X 0.3)(273^4 - 312.7^4)

Q/T = -13.21

Using the higher end of emissivity constant,

Q/T = eoA(T1^4 - T2^4)

Q/T = (0.997)(5.67 X 10^-8)(0.2 X 0.3)(273^4 - 312.7^4)

Q/T = -13.59

From this, we can see that the heat flow due to radiation is greater than the heat flow due to insulation (which is true because although snow has a lot of air, it is still very cold and cannot insulate the soil as much since the soil temperature is much higher). 

References

https://nsidc.org/cryosphere/snow/science/formation.html 

http://www.cnyweather.com/wxsoil.php

http://www.engineeringtoolbox.com/radiation-heat-emissivity-d_432.html

Water Turns Into Steam Immediately In Burning Pan

We see that in the first scenario, it is conduction because heat is being transferred through molecules colliding in the aluminum. In the second scenario, it is convection because the water molecules are moving from one place to another.

Heat lost through the windows!

Today in class we calculated how much heat was lost from windows due to conduction. This made me curious and I wanted to apply it to one of my dad's projects (he builds houses). This particular project is a "bunk house" for a family country home and has a lot of windows. These are the plans for the east side of the bunk house:


(SK Design Group; Pittsfield, MA)

The area for the "F" windows, after converting to meters is 3.2 meters squared, for "H" is 0.84 meters squared, "L" is 1.8 meters squared, and "R" is .94 meters squared. For simplicity, we are assuming the windows are solid glass with thickness .0191 meters.

Plugging these values into the equation for conduction (Q/t=kA ((T1-T2)/l), and adding all of the windows' totals together, I calculated that the east side of the bunk house loses about 16,700 Watts.  Since there are four sides to a house, assuming this is constant, we can say that the house loses roughly 66,800 Watts from windows and glass doors.

These windows aren't actually made of solid glass, however. They are double paned with a 13 mm "airspace" in between the glass panes to trap the heat inside. This prevents both conduction and convection. In the airspace, this particular project has argon gas in it, which is heavier than air and provides better insulation. It also helps soundproof, prevents frost, and can sometimes block UV rays. Some cons are that the gas eventually dissipates (at an unknown rate) and if there isn't a tight seal and the gas leaks out, you will never know. This gas isn't toxic and doesn't have an odor, but you could be losing your insulation without even knowing it.



https://www.thebalance.com/cost-benefits-and-drawbacks-of-argon-gas-windows-844558

Why the Door Knob?

During fire safety trainings, it is common advice to evaluate the conditions of the next room over by feeling the doorknob with the back of your hand. To explore this advice, I wanted to compare how heat is transferred by conduction through wood (the rest of the door) vs. brass (the doorknob).

If we assume that there is a fire in the next room which has heated the air in the room to 500 degrees Fahrenheit (260 degrees Celsius - average fires can be 1000 to 1500 degrees Celsius themselves), then we can model how much heat would be transferred through an area of the door around 0.01 square meters big. The thermal conductivity coefficient of wood is 0.13 W/m*K)We will also assume the room you're in has only been heated slightly by the fire to 30 degrees Celsius, and that the door is 0.035 meters thick (based on 1 3/8 inch door width):

Q/t = kA(T1 - T2)/l = (0.13 W/m*K)*(0.01 m^2)*(260 - 30 degrees Celsius)/(0.035 m) = 8.54 J/s

For the doorknob, even if we assume a slightly smaller cross-sectional area (ex. the back of your hand is larger than the doorknob, so new area is 0.008 m^2) and a longer l (0.20 meters for the entire length of the doorknob on both sides), the heat that passes through the doorknob to your hand is still much greater. The thermal conductivity coefficient is 109 W/m*K:

kA(T1 - T2)/l  = (109 W/m*K)*(0.008 m^2)*(260 - 30 degrees Celsius)/(0.20 m) = 1002.8 J/s

Despite a smaller cross-sectional area and a longer distance between regions of different temperatures, the doorknob would still transfer much more heat from the other room to your hand than the wooden part of the door, and so it is much easier to gain a sense of how hot the other room is using the doorknob.

Source: http://www.farm.net/~mason/materials/thermal_conductivity.html for thermal conductivity coefficients.

Heating Up

In this time of year, dead winter approaches. As it grows colder and colder, the urge to open my window at night with the fan on and the heater going slowly disappears as the window stays shut and the heater starts blasting. However, as we talked about the movement of heat in physics, I grew interested and noticed it all around; in our sweaters, our food, our bodies- right now, it's a very seasonal and on-point topic. While our current heaters blow air into our rooms, electric heaters use radiation to slowly heat up the air in the room. As the thermostat goes up and down these days, I wanted to see how long it would take for an electric heater to heat up an empty small dorm-sized room from about 2C to the heater's comfortable temperature of 23C.

A nifty little aluminum heater. (http://img.hisupplier.com/var/userImages/2008-11/11/hisunrays$011813759(s).jpg)

First of all, since the heater is using radiation, we will be looking at the equation for heat and radiation. To start off, here is the equation for the net heat flow in radiation. The second equation beneath it is changed in order to find time.

Now for our variables; we are assuming that the heater is made from aluminium. To obtain the surface area of the radiator, I assumed a 1mx0.3mx1m sized radiator.

 Now, we must find Q in this instance, as we have each variables besides it. Below is the equation for Q, or heat:

The mass of the air in the room is calculated below. I am assuming a small room that is 6mx5mx3m, (3m tall in consideration of objects present in a room that would normally be 4m tall) resulting in the volume of air to be 120m cubed. Now, as a cubic meter of air is about 1.3 kg, I calculated further to get a mass of 156 kg of air.

Furthermore, taking our change in temperature (23-2=21) and the specific heat of air:

Plugging it in gives us a heat of 3292.38J! Lastly, now that we have Q, we can plug everything back in...

Thus, we see that it takes 162.6s, almost 3 minutes, to fully heat a room to a nice and comfy 23C. So, next time you turn up the heater, don't fiddle with it until it fully heats the room up- you want your room to be heated not too hot, not too cold, but just right!

Please Return Stolen Jacket!

In my EMT training we learned about the different kinds of treatments we would need to perform for conditions that occur due to extreme environmental factors, such as hypothermia. Right now, people, and especially college students who may have altered judgment, need to be extremely careful about dressing appropriately for the weather when going outside. Every year we hear about college students freezing to death after leaving parties in the winter. I decided to look into the net rate of heat flow from a human who is scantily clad vs. the heat flow when wearing a thick down coat on a night like tonight, which will feature a temperature of 20°F (-6.7° C, 266.3 K) at midnight.

For the scantily clad person we need to consider radiation using this equation for the net rate of heat flow:

Q/t= eσA(T₁⁴ –T₂⁴)

Given the A of a person=1.7m², the e_{human skin}= 0.98, and the human body temperature of 310 K

Q= (0.98)(5.67 x 10^-8 W/m²K⁴)(1.7m²)((310K)⁴-(266.3K)⁴)

Q=397 W

When wearing a coat the person is insolated by the air contained within the ~2in (.05m) of down (assume the coat covers the whole body).  

We can find the heat transfer based on conduction:

Q/t= KA((T₁ –T₂)/l)

Given the cross-sectional area of a person is ~0.5m², the thermal conductivity constant of air is 0.024 W/mK, and the distance between the two temperatures is the thickness of the coat

Q=(0.024 W/mK)( 0.5m²)((310K-266.3K)/.05m)

Q=10.4 W

Obviously, the cross sectional area of a person is less than the total surface area used in the calculation of radiation. However, one can still see that much less heat is transferred to the environment when wearing a coat. Jacket thieves run rampant at Colgate, as evidenced by the frequent pleas for the return of coats on the class Facebook pages. We can see here how important coats are for maintaining body heat, and should be more careful to bring our own outerwear and not steal that of others!

References:

http://onlinelibrary.wiley.com/store/10.1256/wea.29.02/asset/2002571103_ftp.pdf?v=1&t=iwidutfm&s=018b03ff369ae002c3a08fb1b7b4b7b8a8ff3287

https://weather.com