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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 6

Lesson 3: Heat capacity and calorimetry

# Heat capacity

Heat capacity is the amount of heat required to change the temperature of a given amount of matter by 1°C. The heat capacity of 1 gram of a substance is called its specific heat capacity (or specific heat), while the heat capacity of 1 mole of a substance is called its molar heat capacity. The amount of heat gained or lost by a sample (q) can be calculated using the equation q = mcΔT, where m is the mass of the sample, c is the specific heat, and ΔT is the temperature change. Created by Jay.

## Want to join the conversation?

• At Jay says that it doesn't rally matter what units you use here, why is that? If I wanted to convert it from Celsius to kelvin would I still add 273 to the temperature? • We're dealing with change in temperatures rather than single temperatures. And a change in a degree Celsius is the same as a change in Kelvin; an increase of 1 degree Celsius is the same as an increase of 1 kelvin. Therefore the delta T (and the specific heat capacity) can use units of Celsius or kelvin.

So for the example at , the ΔT could be 11K or 11 degrees C since a change in either is equivalent.

So if they wanted the initial or final temperature (a single temperature value, not a change), you would have to use a conversion depending which units you wanted.

Hope that helps.
• why does jay write 1.000 g instead of 1 g? • Jay includes those zeros because it reflects greater precision of the instrument he used to measure the mass of the water. In science we want to report as many significant figures as is allowed by our instrumentation, no more no less. So mathematically we could have wrote 1 g and the answer numerically would the same, but we would have sacrificed our precision. In science we have to be conscious of things like the limitations of measurements due to instruments and uncertainty associated with those measurements.

Hope that helps.
• What is q in the equation of heat capacity , it can't be the heat required since the heatrequired is the heat capacity itself
Help • at , why was did he add 75.2J in 1 mol of water to increase temp by 1 degree? is the energy added simply Cm times the number of mols of water? • At -, Jay says that it equals 22K or 22 degrees Celsius. Why is this? Why wouldn't it be -251 degrees Celsius, or something like that? You convert between Kelvin and Celsius by adding or subtracting 273, right? Thanks! • The size of each unit in the Celsius scale is equivalent to each unit in a Kelvin scale. The only difference is that the Kelvin scale is shifted (or vice versa).

If we were calculate T, a temperature, you would be right to assume that we should take this into account. However, we are calculating ΔT. Notice that (x1 + 273) - (x2 + 273) is in fact equal to (x1) - (x2). In other words, the extra "273" in the Kelvin scale is canceled out and the only bit of importance is the actual difference in temperature.

It is worth noting that this would not work for the Fahrenheit scale since each unit in it is smaller.

Hope this helps.
• Can the value of a substance's heat capacity also mean if the temperature of the object decreases by 1º C or 1 K, that amount of heat is transferred to the surroundings? For example, if 1 gram of water decreases its temperature from 15.5ºC to 14.5ºC, does it mean 4.18 J of energy is released in the surroundings?

Also, I'm still having trouble understanding why the heat capacity can exchange ºC or K for the units of temperature. Why is it that a change in degree Celsius is the same as a change in Kelvin (isn't the change scaled by 273 K for 0ºC)?
(1 vote) • Yeah your first assumption is correct. If you work out the math then the heat associated with a negative change in temperature gives the heat a negative sign. It'll have the same magnitude as the same increase in temperature, just different signs. So yeah it can be thought of as the amount of energy required to raise the temperature, and also the amount of energy lost by a decrease in temperature.

For the second question, we can exchange the temperature part of heat capacity's unit because the equation, Q = (M)(C)(ΔT), uses the change in temperature as opposed to a single temperature value. If we're using a single temperature, then it does matter which temperature scale and unit we're using. However here with a change in temperature it doesn't matter since the Celsius and Kelvin scales are defined such that a change in one scale is equivalent to the other. Mathematically you can imagine this as the two scales being lines having the same slopes, only differing in their y-intercepts. So since a change in Celsius is the same as a change in Kelvin, we can use either unit.

For example, individual temperatures like 25°C or 28°C are different from 298K and 301K, but the ΔT is the same for both scales (°C: 28-25 = 3 and K: 301-298 = 3).

Hope that helps.
• Shouldn't the change in mass also be taken into account when calculating heat capacity? • but beacuse of the why is the because happening?
(1 vote) • thank you sir and guys, really good video, i was trying to figure out how much heat required for my work
(1 vote) • @ in the video he just says "when we do the math" and doesn't really go into or explain the math, i dont understand how from: 1.0*10^2=10.0g(0.90J/g-K) could produce the value 11 for the change in temperature.
(1 vote) • It’s an algebra problem where we’re solving for ΔT in the heat capacity equation.

So, q = mCΔT, given equation
q/(mC) = ΔT, divide by m and C for both sides of the equation. So ΔT is solved for then we substitute the values in to get a numerical answer.

(1.0 x 10^(2))/(10.0*0.90) = 11.111… or 11 when rounded for sig figs.

Hope that helps.
(1 vote)

## Video transcript

- [Instructor] The heat capacity of an object is the amount of heat necessary to raise the temperature of the object by one degree Celsius or one Kelvin. The specific heat capacity, which is often just called specific heat is the heat capacity of one gram of a substance while the molar heat capacity is the heat capacity for one mole of a substance. We symbolize specific heat with a capital C and a subscript s for specific, and molar heat capacity is symbolized by capital C with a subscript m. First let's look at specific heat. The specific heat of water is equal to 4.18 joules per gram degrees Celsius. And what this means is if we have one gram of liquid water, and let's say the initial temperature is 14.5 degrees Celsius, it takes positive 4.18 joules of energy to increase the temperature of that one gram of water by one degree Celsius. Therefore the final temperature of the water would be 15.5 degrees Celsius after we add 4.18 joules. Next let's calculate the molar heat capacity of water from the specific heat. If we multiply the specific heat of water by the molar mass of water which is 18.0 grams per mole, the grams will cancel out and that gives us 75.2 joules per mole degree Celsius. And so this is the molar heat capacity of water. Let's say we had 18.0 grams of water. If we divide by the molar mass of water which is 18.0 grams per mole, the grams cancel and that gives us one mole of liquid water. So one mole of H2O. Using the molar heat capacity of water, it would take positive 75.2 joules of energy to increase the temperature of that 18.0 grams of water by one degree Celsius. Next let's calculate how much heat is necessary to warm 250 grams of water from an initial temperature of 22 degrees Celsius to a final temperature of 98 degrees Celsius. Using the units for specific heat, which are joules per gram degree Celsius. We can rewrite the specific heat is equal to joules is the quantity of heat that's transferred. So we could just write q for that. Grams is the mass of the substance and degree Celsius is talking about the change in temperature delta T. So if we multiply both sides by m delta T, we arrive at the following equation which is q is equal to mC delta T. And we can use this equation to calculate the heat transferred for different substances with different specific heats. However, right now we're only interested in our liquid water and how much heat it takes to increase the temperature of our water from 22 degrees Celsius to a final temperature of 98 degrees Celsius. To find the change in temperature, that's equal to the final temperature minus the initial temperature which would be 98 degrees Celsius minus 22 which is equal to 76 degrees Celsius. Next, we can plug everything into our equation. Q is what we're trying to find. M is the mass of the substance, which is 250 grams. C is the specific heat of water which is 4.18 joules per gram degrees Celsius, and delta T we've just found is 76 degrees Celsius. So let's plug everything into our equation. Q would be equal to, the mass is 250 grams. The specific heat of water is 4.18 joules per gram degree Celsius. And the change of temperature is 76 degrees Celsius. So looking at that, we can see that grams will cancel out and degrees Celsius will cancel out and give us, q is equal to 79,420 joules or to two significant figures, q is equal to 7.9 times 10 to the fourth joules. So 7.9 times 10 to the fourth joules of energy has to be transferred to the water to increase the temperature of the water from 22 degrees Celsius to 98 degrees Celsius. The specific heat can vary slightly with temperature. So the temperature is often specified when you're looking at a table for specific heats. For example, in the left column we have different substances, on the right column we have their specific heats at 298 Kelvin. So we could use for our units for specific heat joules per gram degrees Celsius, or we could use joules per gram Kelvin. For liquid water, the specific heat is 4.18 at 298 Kelvin. For aluminum, solid aluminum, the specific heat is 0.90. And for solid iron the specific heat is 0.45 joules per gram Kelvin. Let's compare the two metals on our table here. Let's compare a solid aluminum and solid iron. So we're gonna add 1.0 times 10 to the second joules of energy to both metals and see what happens in terms of change in temperature. First, let's do the calculation for aluminum. We're doing Q is equal to mC delta T and we're adding 1.0 times 10 to the second joules. And let's say we had 10 grams of both of our metals. So this would be 10.0 grams of aluminum. And then we multiply that by this specific heat of aluminum, which is 0.90. So 0.90 joules per gram Kelvin times delta T. When we do the math for this, the joules will cancel out, the grams will cancel out and we would find that delta T would be equal to 11 Kelvin or 11 degrees Celsius. It doesn't really matter which units you're using here for the specific heat. Next let's do the same calculation for iron. So we're adding the same amount of heat. So 1.0 times 10 to the second joules of energy. So we can plug that in, 1.0 times 10 to the second joules. We're dealing with the same mass so we have 10.0 grams of iron but this time we're using these specific heat of iron which is 0.45 joules per gram Kelvin times delta T. So once again, joules cancels out, grams cancels out and we get that delta T is equal to 22 Kelvin, or 22 degrees Celsius. What we can learn from doing these two calculations is we had the same amount of heat added to our two substances with the mass of the two substances was the same, the difference was their specific heats. So iron has a lower specific heat than aluminum. And since iron has a lower specific heat, it's easier to change the temperature of the iron. So the lower the value for the specific heat, the higher the change in the temperature, or you could also say the higher the value for the specific heat, the smaller the change in the temperature, and going back to our chart, water, liquid water has a relatively high specific heat which means the temperature of water is relatively resistant to change.