Wednesday, November 11, 2015

Heat Capacity (for people who prefer formulas and algebra)

Heat capacity problems are always interesting because they are one of the types of science problems that you can often work through with just a little bit of insight into what heat capacity is, even if you don't know the "correct" way to do the problem. Heat capacity is the energy required to change the temperature of a substance. If you can keep track of energy, amount of the substance, and the temperatures before and after the change, you just might be able to cobble together the quantities and arrive at the right answer.

But I know, sometimes you just want to memorize a mathematical formula and plug numbers in. If that's what you're looking for, then this is just for you. Heat capacity problems can pretty much all be solved using the formula:
(Energy transferred) = (Heat capacity) x (Amount of substance) x [(Final temperature) - (Initial temperature)]
Or, if we want to make that shorter:
E = (Cp) x (g) x (Tfinal - Tinitial)
Let's plug in information from 2 different problems:

1. A 250.0g sample of water (Cp = 1 calorie/(g)(°C)) is heated from 14.3°C to 27.4°C. How many calories of heat energy have been transferred?
Plugging in to the formula:
E = (1 calorie/(g)(°C)) (250.0g) (27.4°C - 14.3°C) = 3275calories

2. A 400.0g sample of water is initially at 16.8°C. If 5000 calories of energy is added to the water, what is the final temperature?
Plugging in again:
5000 calories = (1 calorie/(g)(°C)) (400.0g) (Tfinal - 16.8°C)
Same thing, but now we have to do a little algebra to solve for Tfinal, and we get a final temperature of 29.3°C.

If you prefer a more descriptive solution to heat capacity problems, take a look at

Thursday, November 5, 2015

Heat Capacity

There have been a few questions about heat capacity…

Heat capacity is a measure of the amount of heat required to change the temperature of a given amount of a substance by some amount. A common unit for heat capacity is "calories / (gram)(°C)". For water, heat capacity is 1 calorie / (gram)(°C), so if I have 1 gram of water and I want to increase its temperature by 1°C, I have to add 1 calorie of energy to the water. What if I have more than 1 gram or I want to increase the temperature by more than 1°C? Multiply!

The reverse of this problem is really the same problem, it just requires some different math. What if I have 18.00mL of water that is initially at 12.6°C and I add 49calories of energy to that water? First part… the density of pure water is 1 g/mL so 18.00mL of water has a mass of 18.00g. Now, if the heat capacity of water is 1 calorie / (gram)(°C), and we have 18.00g of water, we can again multiply to get:
{1 calorie / (gram)(°C)} x 18.00g = 18.00 calories per °C
So for every 18.00 calories of energy we add to this specific sample, we will increase the temperature by 1°C. We are adding 49 calories to this specific sample so:
49 calories / 18.00 calories per °C = 2.7°C
This is how much the temperature changes when we add this amount of energy. Since the sample was initially at 12.6°C and we added energy, the new final temperature must be 2.7°C higher than the initial temperature, 12.6°C + 2.7°C = 15.3°C.

There are some assumptions in this description (like the density of water) that simplify the problem… if you want to get the absolutely perfectly correct answer, you'd have to take some of those assumption into account, but this is close enough for our purposes.

Monday, November 2, 2015

Microwave Heating Experiment (Fall 2015)

There are quite a few questions about the Microwave Heating experiment (procedure here), so let me try to address them all in one place.

The most common and persistent questions are related to the data analysis. The hands-on procedure and collection of the data seems to be going well, so let's not dwell on that part of the experiment. Once you have collected your data, you'll have a sea of numbers that you, as a scientific investigator, will have to process and interpret so you can get some useful information out of your experiment. The data you are collecting will look something like this:
The procedure tells you to prepare a graph of this data. Whenever you are collecting data (or just observing something interesting in your daily life) where one variable is under your control (in the case of this specific example, time), and another variable is simply a quantity you are measuring (in this case, temperature), it is ALWAYS interesting to consider what the graph of that information looks like. In this case, the graph of the above data looks like:
Hmm, interesting, although in this case not too surprising: as more time in the microwave passes, the sample gets hotter. But how much hotter? And how fast does it get hotter? If I want to heat a cup of water to make tea, should I put it in the microwave for 30 seconds or 5 minutes? The rate of heating is an interesting and important piece of information we can extract from this data!

How do we calculate a rate? Rate is the change in some observable quantity divided by the change in time. If you can eat a cheeseburger in 5 minutes, your rate of eating cheeseburgers is:
Change in the number of cheeseburgers / Change in time
1 cheeseburger / 5 minutes = 0.2 cheeseburgers per minute
In this experiment, you weren't eating cheeseburgers (well, OK, I guess you might have been eating cheeseburgers while you were watching water heat up…), but you were observing the change in temperature as time passed, so we should be able to calculate a rate in a similar way. Looking back at the data above, we can pick any two points and look at the change in temperature divided by the change in time. For example, from 90 seconds to 135 seconds the temperature changed from 116.04°C to 184.70°C, so the rate over that time period was:
(184.70°C - 116.04°C) / (135 seconds - 90 seconds)
(68.66°C) / (45 seconds)
1.5°C per second
We could pick any two points out of our data and calculate a rate, and they would all be pretty similar, but because there is some variability in our experiment they would not all be identical. But wait, if that's all we are going to do, then why did we make a graph? Graphs are pretty, but making a graph for the sake of making a graph doesn't seem like a great use of your time. Can we use the graph to determine the rate of heating? Hmm, change in temperature… the vertical axis (y-axis) is temperature, so that one should track changes in temperature… and the horizontal axis (x-axis) is time… WAIT! The data points in the graph look pretty close to linear, and the slope of a line is "rise over run"… if "rise" is the change in the y-axis variable and "run" is the change in the x-axis variable, then we should be able to get the slope from the graph, and the slope should be the rate of heating! If we draw a line that looks like it fits the data pretty well, we get:
From the line (we're not looking at specific points now), it looks like over the time period from 0 seconds to 300 seconds the temperature rises from about 5°C to about 390°C. So the rate based on the fit line is:
(390°C) / (300 seconds)
1.3°C per second
By using a fit line, we can even out some of the variability associated with any two points we might choose and we should get a more reliable answer. If we look at the rate for any two adjacent points (any 15 second period in our experiment) and calculate the rate using only those two points, we would get answers as low as 0.75°C per second and as high as 1.9°C per second, so using the fit line seems like a pretty good way to get a reliable single answer for the experiment.

Good luck on your data collection and analysis.

Tuesday, August 25, 2015

Fall 2015 - Welcome to class!

Fall 2015 semester has begun! Welcome to BCBT 100 - The Science of Cooking. Together we will explore the scientific foundations that humans (and pre-humans...) have used for millenia to produce and transform food. If you have ever prepared or even just eaten food, you are a scientist!

Let's have a great semester!