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

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