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Heat Capacity And Calorimetry - Example 2

Heat capacity is the amount of heat required to change the temperature of a mass unit by one degree. The symbol for heat capacity is "C". The specific heat capacity of a substance is the heat capacity per unit mass. The heat capacity of a substance varies with temperature, and certain materials have a very large heat capacity. The specific heat capacity, also called the specific heat, is the heat capacity per unit mass of a substance. The SI unit for specific heat capacity is joules per kilogram kelvin, although the calorie (cal) and the kilogram calorie are still in wide use.


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Video Transcript

Welcome to our second example video. Looking at specific heats and Kalorama Tree this video. We're going to consider some water telling ourselves that we have 2 kg of water that are at that are frozen, and we want to find out how much energy is it going to take to turn these 2 kg of water I e. Ice into water vapor. Okay, so the first thing we need to be able to raise the temperature of the ice up to the melting temperature, which is zero degrees Celsius If we say we start at negative 10 degrees Celsius ice, that means that the heat is going to take, so we'll call this Q one. The heat is going to require to raise a 10 degrees Celsius is going to be the mass 2 kg multiplied by the specific heat of ice, which is 2090 Jules per kilogram. Kelvin multiplied by the change in temperature, which in this case is going to be 10 now, remember, because this is a change in temperature. 10 kelvin is the same as 10 degrees Celsius. Next, we need to deal with the latent heat of fusion. That is, we're going to go from a solid to a liquid were melting it. So then cue to the next step will be equal to the Mass 2 kg times latent heat of fusion, which is 3.33 times 10 to the five Jules per kilogram. Hopefully, you can see here we're already going to have a very large number of jewels required to do this. Next, we need to heat it up to our vaporization temperature, which is 100 degrees Celsius. So that's going to be 2 kg of liquid water multiplied by it's specific heat, which is 4190 jewels per kilogram. Kelvin, all multiplied by the change in temperature, which is 100. And then we have to vaporize. That's that's going to be Q four is equal to 2 kg of still liquid water multiplied by the latent heat of vaporization, which is 22.6 times 10 to the five Jules per kilogram. So that is a lot of heat that it is going to require in order to get this to these 2 kg to go all the way from being frozen, being ice all the way up to being water vapor. Now, um, this amount of heat kind of explains why it is so expensive to heat things industrially, if you aren't familiar with the industry, requires enormous temperatures because they generally require all sorts of phase changes. It's one of the ways that they tend to purify materials is by changing the face, Um, and so it costs a lot of heat to get through that latent heat of fusion and latent heat of vaporization.