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Heat Capacity And Calorimetry - Example 4

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 fourth example video, looking at specific heat and also at Kalorama Tree. In this video, though, we're going to be looking at a diabetic compression. We talked about the Adia Batic process and what it takes to calculate what's happening there. Because remember, this is a situation where Q he transferred is equal to zero. Now the way they accomplished this is by either insulating the system where the automatic process is happening or they just make it happen so fast that there is no opportunity for heat to be transferred. S O. For example, if you're thinking about an internal combustion engine, you have the piston going up in sound in the cylinder so fast that there's just no time for heat exchange. So essentially is an ATI a Batic process. Incidentally, the same thing could be true for Kalorama Tree. When we're looking at our that of liquid with stuff inside, we have to insulate this ah lot. Otherwise we'll lose heat through the process of Kalorama tree itself with the external world like with the air up above the water Okay, returning to the problem at hand. Though what we have is we have a cylinder with an initial volume of 400 cubic centimeters or 400 cc, and it's going to squish down what's inside, which is air mostly down to 50 ccs. It starts at a pressure of one atmosphere at a temperature that's a little of normal 30 degrees Celsius. And we want to find out what are the final pressure, the final temperature and what's the work going to be in order to solve for all this? Um, remember that when it comes to automatic processes, we had a very important constant, which was gamma for air real use 1.4, which is for a dye atomic gas, because air is mostly, uh, nitrogen to an oxygen to so that's end to a no to make up the majority of it. And the Mueller specific heat for constant volume, which we can use to solve for the work, is simply 20.8 jewels promote Calvin, which is the same as for nitrogen, which is the majority of what air is composed of. Okay, so what can we do here? Well, we know we have p one V one to the gamma is equal to p final times V final to the Gamma. And over here we know the initial volume. We know the final volume. We know the initial pressure, which means the final pressure is going to be equal to the initial pressure. P one multiplied by V one over V final to the gamma power so you can go ahead and plug that in noticed. Since we're taking a ratio of the volumes, there's no need to convert this into cubic meters because the units will cancel out. Anyways. It's just going to be 400 divided by 50 and that's it. So, um, we need to remember to take it to the power of 1.4 a. Swell. So that's our final pressure. So for two final temperature, we can use a similar equation which looks like t not be not to the gamma minus one. It's going to be equal to T final V final to the gamma minus one, and you can find T final with a really similarly looking equation where we have V not over v final to the Gamma minus one. So really, all that's changed are the variable names and the factor to which we are raising Gamma. Finally, for the work, remember that work for an automatic process is equal to N C V Delta T Delta T is gonna be t final minus t initial. So we have that we have C V is given by nitrogen, and then we need to find the moles. The number of moles. Well, luckily, we also have the ideal gas law, she tells us. PV equals N. R T. You can either pick the final state or the initial state. Either way, you're going to get P is equal Thio and is equal to P not being not over our peanut. Or it could be equal to P final. The final over R B final. You'll get the same number of moles and you can plug that in to calculate your work.