Video transcript
- [Instructor] The heat capacityof an object is the amount of heat necessary to raise the temperature of the object by one degreeCelsius or one Kelvin. The specific heat capacity, which is often just called specificheat is the heat capacity of one gram of a substance while the molar heatcapacity is the heat capacity for one mole of a substance. We symbolize specificheat with a capital C and a subscript s for specific, and molar heat capacityis 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 wehave 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 waterby one degree Celsius. Therefore the finaltemperature of the water would be 15.5 degrees Celsiusafter we add 4.18 joules. Next let's calculatethe 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 outand that gives us 75.2 joules per mole degree Celsius. And so this is the molarheat 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 permole, the grams cancel and that gives us onemole 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 waterby one degree Celsius. Next let's calculate howmuch heat is necessary to warm 250 grams of waterfrom an initial temperature of 22 degrees Celsiusto 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 quantityof 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 equationto calculate the heat transferred for different substances with different specific heats. However, right now we're onlyinterested in our liquid water and how much heat it takesto increase the temperature of our water from 22 degreesCelsius 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 equalto 76 degrees Celsius. Next, we can plug everythinginto our equation. Q is what we're trying to find. M is the mass of thesubstance, which is 250 grams. C is the specific heat of water which is 4.18 joules pergram degrees Celsius, and delta T we've justfound is 76 degrees Celsius. So let's plug everythinginto 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 temperatureis 76 degrees Celsius. So looking at that, we cansee that grams will cancel out and degrees Celsius willcancel out and give us, q is equal to 79,420 joulesor to two significant figures, q is equal to 7.9 times10 to the fourth joules. So 7.9 times 10 to thefourth joules of energy has to be transferred to the water to increase the temperature of the water from 22 degrees Celsiusto 98 degrees Celsius. The specific heat can varyslightly with temperature. So the temperature is oftenspecified when you're looking at a table for specific heats. For example, in the left column we have different substances, on the right column wehave their specific heats at 298 Kelvin. So we could use for our units for specific heat joulesper gram degrees Celsius, or we could use joules per gram Kelvin. For liquid water, the specificheat is 4.18 at 298 Kelvin. For aluminum, solid aluminum,the specific heat is 0.90. And for solid iron thespecific heat is 0.45 joules per gram Kelvin. Let's compare the twometals on our table here. Let's compare a solidaluminum and solid iron. So we're gonna add 1.0 times 10 to the second joules ofenergy to both metals and see what happens in termsof change in temperature. First, let's do thecalculation for aluminum. We're doing Q is equal to mC delta T and we're adding 1.0 times10 to the second joules. And let's say we had 10grams of both of our metals. So this would be 10.0 grams of aluminum. And then we multiply that bythis specific heat of aluminum, which is 0.90. So 0.90 joules per gramKelvin times delta T. When we do the math for this, the joules will cancel out,the grams will cancel out and we would find thatdelta T would be equal to 11 Kelvin or 11 degrees Celsius. It doesn't really matterwhich units you're using here for the specific heat. Next let's do the samecalculation for iron. So we're adding the same amount of heat. So 1.0 times 10 to thesecond 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 usingthese specific heat of iron which is 0.45 joules pergram Kelvin times delta T. So once again, joules cancelsout, grams cancels out and we get that delta Tis equal to 22 Kelvin, or 22 degrees Celsius. What we can learn fromdoing 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 differencewas their specific heats. So iron has a lowerspecific heat than aluminum. And since iron has a lower specific heat, it's easier to change thetemperature of the iron. So the lower the valuefor 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 arelatively high specific heat which means the temperature of water is relatively resistant to change.
