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 = (C_p) x (g) x (T_final - T_initial)

Let's plug in information from 2 different problems:

1. A 250.0g sample of water (C_p = 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) (T_final - 16.8°C)

Same thing, but now we have to do a little algebra to solve for T_final, 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 http://scienceofcooking100.blogspot.com/2015/11/heat-capacity.html

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.