Energy

Thermodynamics

Question:

How would the energy needed to change 10 grams of water by 1 degree Celsius, be compared to changing 100 grams of water by 1 degree Celsius? This question has got to do with specific heat capacity. Thank you.

Answer:

This question is from the area of thermodynamics:

First off the specific heat capacity (or more commonly specific heat) of a material is a measure of the amount of ‘heat’ required to raise the temperature, of a sample, by a unit of temperature. Since you are increasing your sample size by ten times (i.e. 10 to 100) intuition should tell you that the heat required will increase and as expected this is the case! To keep things as simple as possible let us assume that both samples are initially at 20 degrees and both at the same pressure. We are doing this to limit the differences between the two set-ups and minimize the complexity of the problem. This way the answer and analysis can focus on what you describe as the significant change in your experiment, namely, the change in mass. Under these circumstances we can use the following formula:

Q (heat required) = m (mass) * dT (difference or change in temperature) * c (specific heat constant)

The specific heat constant denoted as ‘c’ has units of: J/(g*K). The ‘J’ represents units of energy known as Joules, ‘g’ represents mass in grams and ‘K’ represents our temperature scale in kelvins. Since your question uses Celsius it should be noted that a change of 1 K is equal to a change of 1 degree Celsius.

In this case, for both samples the temperature change and the specific heat constant are the same so after simplifying all we are left with is the change in mass. Since our sample increases by a factor of 10 the amount of ‘heat’ required will also increase by a factor of 10. Mathematically we can produce this ratio as follows:

Q(sample 2 )/ Q(sample 1) =
[m(sample2) * dT(sample2) * c] / [m(sample1) * dT(sample1) * c]

Since dT(sample2) and dT(sample1) are equal (i.e. 1 in this case) they, along with the ‘c’ term which is common to both top and bottom, cancel and you are left with:

Q2 / Q1 = m2 / m1 = 100/10 = 10

* If we were to change the experiment so that the samples were subjected to different conditions (examples of these changes could include holding the samples at different pressures or selecting a change in the temperature that would cause one of the samples to undergo a phase change) our end answer would change. *

- Eric Chisholm

 


 

Wind Power

Question:

I’m just wondering is it possible to make a car engine that is powered by the motion of wind?

Answer:

At this time, as far as I’m aware, the only viable way for a car engine to be powered by wind is by using an electric vehicle and a wind turbine.  The wind energy would be harvested with a stationary turbine and then used to power the electric vehicle.  Currently, the most readily available electric vehicles are plug-in hybrids and battery electric.  Hydrogen fuel cell vehicles are also electric vehicles and not far behind the competition. 

For more information, please visit the following web pages on electric vehicles, hydrogen and fuel cells and renewable power.

http://oee.nrcan.gc.ca/transportation/alternative-fuels/index.cfm?attr=8

http://www.h2fcprogress.collaboration.gc.ca/scie/index-eng.htm

http://canmetenergy-canmetenergie.nrcan-rncan.gc.ca/eng/renewables/wind_energy.html

 


 

Vehicles

Question:

My questions are based on the following assumption: The fact that passenger cars share the road with heavy vehicles (transportation of goods) requires standards organizations to adopt stringent safety requirements that are costly in terms of resources. This translates into relatively heavy and complex cars that consume a lot of energy over their life cycles. Assuming it were possible to reorganize the road network to minimize the sharing of the road by heavy vehicles and cars, it would be feasible to allow individuals to use very light (comparable to the weight of a motorcycle) and therefore much more energy-efficient cars. If the necessary changes to standards included a plan to extend the life of these cars, thereby inhibiting planned obsolescence, which consumes more natural resources and energy, it is likely reasonable to assume that the natural life of these cars could be doubled.

Is the potential energy economy quantifiable (on a country-wide scale and for the entire life cycle of the vehicle)? Is this topic already being studied in a Canadian research centre or university? Could these ultra-light vehicles technically be powered by an electricity distribution network running alongside roadways, using the same operating principles as a streetcar system?

Answer:

A model could certainly be developed for quantifying energy use (or economy) for such a type of vehicle. Given certain parameters on the weight reductions incurred from these lighter designs, the GHGenius (http://www.ghgenius.ca/) tool developed by NRCan would probably be a good basis for conducting a life cycle analysis. Assuming all personal motor vehicles in Canada were made to follow suit, applying it on a country-wide scale is just a question of modeling an ‘average’ car in GHGenius and then multiplying by the number of cars in Canada (this would give an approximate solution, the accuracy of which would depend on the assumptions made).

Ultra-light vehicles could certainly be powered by an electricity distribution network. The technology for this type of power distribution is well-defined and exists. The better question to ask here is whether this is feasible, economical, and practical. Running lines across every road and highway in Canada would be prohibitively costly to both implement and maintain. It makes sense when the movement of traffic is on a defined path (like a tram line or bus route) but not necessarily as a widespread system for all Canadians. The problem with such a system is that you can only go where there are lines to power the vehicle; the vehicle is dependent on the infrastructure of both the roads and the power source.

The problem is an environmental, economic, and engineering based problem, so there are plenty of places where this research might be taking place. Here are some contacts for further information:

NSERC research chairs that are conducting automotive light-weighting research:

http://www.nserc-crsng.gc.ca/Chairholders-TitulairesDeChaire/Chairholder-Titulaire_eng.asp?pid=394
http://www.nserc-crsng.gc.ca/Chairholders-TitulairesDeChaire/Chairholder-Titulaire_eng.asp?pid=189

University of Waterloo Centre for Automotive Research: http://watcar.uwaterloo.ca/

 


 

Ionization Energy of Air

Question:

What’s the laser frequency required to ionize air and what’s the ionization energy of air?

Answer:

It is more appropriate to think of air as a mixture of different gases, each one having a different ionization energy. That said, one has to ask oneself which gas they wish to ionize. Let’s look at nitrogen since “air” contains about 78% of it.

Nitrogen has an ionization energy of 1402.3 kJ/mol (amount of energy required to remove 1 mol of electrons from 1 mol of nitrogen gas) or

1402 kJ/mol X 1000 J/kJ / 6.022 X 10^23 = 2.328X10^-18 Joules per atom

Therefore, the wavelength of the laser beam must have at least this much energy to ionize nitrogen. To find the wavelength, we use

E = h*c/lambda

where E is the required ionization energy (in our case 2.328×10^-18), h is Planck’s constant in J*s (http://en.wikipedia.org/wiki/Planck_constant), c is the speed of

light (2.98×10^8 m/s) and lambda is the wavelength of the laser beam.

Plugging in these values and solving for lambda, we get a wavelength of approximately 85nm (far ultra-violet light). Using

f = c / lambda

where f is the frequency, we get a frequency of 3.5×10^15 Hz. It’s a good thing nitrogen requires far ultra-violet light to ionize otherwise sunburns would be far worse!

- Kevin Shortt

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