If you’re new to Python, go to the menu bar and click Runtime > Run all.
Then wait for around 20 minutes while this script scrapes data from the VCAA and generates an interactive scatterplot for you. When it’s done, there will be some interesting data files available for download from the file explorer on the left of the screen.
You’ll notice some interesting findings in the scatterplot, including the fact that boys outperform girls in biology, and girls outperform girls in physics! Girls outperform boys in 15 of the 20 most popular VCE subjects with the only exceptions being Chemistry, Biology (only slightly) and all three mathematics subjects.
Feel free to modify this code and repost it. There are some other interesting insights you could glean from the dataset. Enjoy!
In this analysis, I defined “high-achieving students” as those who achieve at least 2 study scores ≥40. I then compared this with enrolment data to see how their subject choices differed from that of all students (from VCAA statistics).
Choosing these subjects doesn’t guarantee you a high grade. But it does provide some interesting insight into the patterns of high-achieving students, who are more likely to have chosen Specialist Maths, Latin, Chemistry, Global Politics, Physics and Literature.
Today, we’re going to answer the following question:
When 200 grams of ice is added to a bucket containing 1.00 litre of hot water, what’s the final temperature of the water?
To answer the question, we’re going to need to make some assumptions. We’ll take 1.000 litre of pure water at 80.00°C and add 200.0 g of ice (at -10.00°C) to it. What’s the final temperature of the water?
Part 1: Heat transfer method
The following equation can calculate the temperature at thermal equilibrium of any number of objects in thermal contact.
I love this equation because it’s several lines of maths shorter than the version taught in school. With this equation, you don’t even need to convert the temperatures into kelvin. Celsius works just fine.
Let’s set up the equation so that the addition series contains the variables in the question.
Now, let’s substitute the gives values into the equation. The specific heat capacity of water is 4200 J kg-1 K-1, and that of ice is 2100 J kg-1 K-1.
Great! Adding 200.0 g of ice to 1.000 L of water decreases the temperature from 80.00°C to 71.80°C.
But we’ve forgotten something. The ice will melt as soon as it hits the hot water. Since melting is an endothermic process, heat energy from the water will actually be absorbed, thus reducing the final temperature even further.
Part 2: Let’s take into account the fact that the ice melts!
Remember our formula from part 1.
The amount of energy required to melt ice can be calculated using the latent heat equation:
Removing that amount of heat energy from the system results in the following equation:
Great! Now, we’ve calculated that the final temperature of the water would be 57.36°C after the addition of the ice. That’s equal to 330.5 kelvin, which will be useful later.
However, we’ve forgotten to take something else into account: how much heat will be lost as radiation from the surface of the bucket?
Part 3: What’s the rate of heat loss from the bucket by radiation?
The rate of heat lost by radiation can be calculated by using the Stefan-Boltzmann equation, below.
P is the rate at which heat energy is radiated from the surface of the bucket in watts. Emissivity, e, of water is 0.95, and the surface area, A, should be around 0.0707 m2 for a one-litre bucket. Calculation of A is shown below. Assuming that the radius of the surface of the bucket is 6cm:
Plugging that value into the equation, we can find P. We’ll assume that the experiment is being conducted at room temperature and the temperature of the surroundings is 20.00°C (29.03 K).
This means that 2.928 joules of energy are emitted from the surface of the bucket every second. Ten minutes later, the bucket would have lost 1756.8 joules of energy due to radiation from the surface. But what about emission of radiation from the sides of the bucket?
Let’s say that our bucket is made from highly polished aluminium (which has emissivity 0.035) and it holds exactly 1.2 litres of water. We need to calculate the dimensions of the bucket.
Assuming it has straight sides (i.e. it’s a cylinder), the bucket had volume equal to the following formula:
The surface area of our bucket (excluding the open surface at the top) is:
The rate of energy radiation from the sides would therefore be:
It’s interesting to note how very little radiation is emitted from the shiny aluminium bucket, while lots more radiation is emitted from the surface of the water. This is because relatively ‘dark’ water has a much higher emissivity than shiny aluminium. Total emission from the bucket is therefore:
After ten minutes, the bucket would have lost the following amount of energy:
Let’s factor this amount of energy loss into our final temperature equation.
Not much energy is lost via radiation! Finally, let’s find the peak wavelength of the radiation emitted by the object using Wien’s law.
Part 4: What’s the wavelength of the radiation being emitted by the bucket?
Here’s Wien’s law from Unit 1 Physics…
The radiation emitted from the resulting bucket of water lies firmly in the infra-red part of the electromagnetic spectrum. The bucket would be clearly visible on an infra-red camera!
Next week, we’ll begin a new a Chemistry-themed project called Periodic Table Smoothie. More next week.
Inspired by the formula booklets used by VCE Physics and VCE Maths Methods, here’s an 8-page Chemistry formula booklet you can use for your Year 11 and 12 Chemistry assignments. This custom-made booklet is a a collection of reliable formulae that I have been using to answer VCE Chemistry questions while teaching and tutoring.
There are 76 formulae in total, at least 10 of which are original. Orders are shipped on A3 paper, stapled along the spine and folded to an A4-sized booklet that resembles the VCAA Data Booklet.
Orders from schools, students and tutors are all welcome. Price includes free international delivery and 10% voucher for the T-shirt store.
Significant figures tell you how accurately a number is known. This invariably depends on the precision of your instruments.
To illustrate this, use a pencil and a ruler to draw a square with sides of 8.109435 cm in length. Now, calculate the area of the square that you’ve drawn.
A ruler can only measure length to within ±0.1 cm. Our square therefore has sides 8.1 cm in length (not 8.109435 cm) because our measurements are limited by the accuracy of the ruler. The area of our square is therefore 8.1×8.1=66 cm², not 65.762936019225 cm², because there was no way to measure all of those decimal places precisely using a ruler.
Accurately-known digits are known as significant digits. All other digits are described as not significant. We must always round our final answer (not the intermediate steps) to the correct number of significant digits by following the six rules below.
1. Numbers without a decimal point
First non-zero digit is significant
Last non-zero digit is significant
All digits in-between are significant
45 is to 2 significant figures (s.f.)
1,240 (3 s.f.)
68,686,000 (5 s.f.)
2. Numbers with a decimal point
First non-zero digit is significant
All digits afterwards are significant
1.2 (2 s.f.)
6.810 (4 s.f.)
900,001 (6 s.f.)
3. Scientific notation
Scientific notation is a way of writing numbers in the form:
a × 10b where 1 ≤ a < 10.
Count the number of significant figures in a to find the number of significant figures in the number (a × 10b).
5.56 × 103 is to 3 significant figures
2.012 × 10-4 is to 4 significant figures
Some unit conversions are exact and are said to have an unlimited number of significant figures.
1 minute = 60.0000000000… seconds (infinite s.f.)
1 metre = 100.0000000000… metres (infinite s.f.)
Temperatures usually have 3 (sometimes 4) significant figures when converted into Kelvin!
10°C = 283 K (3 significant figures)
100°C = 373 K (3 significant figures)
4000°C = 4273 K (4 significant figures)
5. Addition and subtraction
Rule: Always round your final answer (not any intermediate answers!) to the smallest number of decimal places.
441 + 65.42 = 506 (use zero decimal places)
200.1 – 144.2456 = 55.9 (use 1 decimal place)
6. Multiplication and division
Rule: Always round your answer so it has the same number of significant figures as the input value with the smallest number of significant figures.
481.56 × 14.5 = 6980 (use only 3 s.f.)
7800 ÷ 41.1 = 190 (use only 2 s.f.)
Remember to round your ANSWER (not the intermediate steps) to the correct number of significant figures.
Questions? Comments? Still confused? Leave a message in the comments below. I’ve tried to make sig figs as simple as I can in this post.