About this video
This video defines heat capacity, specific heat (per gram) and molar heat capacity (per mole), then derives and uses the equation q = mc(Delta T) to find the heat needed to warm water. It finishes by comparing the specific heats of aluminum and iron, showing that a lower specific heat gives a larger temperature change for the same heat added.
You'll be able to
- distinguish specific heat from molar heat capacity.
- use q = mc(Delta T) to calculate heat transferred.
- calculate a change in temperature from mass, specific heat, and heat.
- explain how specific heat affects how much a substance's temperature changes.
Key words
- heat capacity
- The heat needed to raise an object's temperature by one degree Celsius or one Kelvin.
- specific heat (Cs)
- The heat capacity of one gram of a substance.
- molar heat capacity (Cm)
- The heat capacity of one mole of a substance.
- q = mc(Delta T)
- The equation linking heat q to mass, specific heat, and temperature change.
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.
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