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Solving Complex Calorimetry Problems involving changes of state and temperature

Added Apr 05, 2019 •

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This video describes how to solve a calorimetry problem in which ice is heated through phase changes from solid to liquid to steam.

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fish videos going to take it through a complex Colorado tree problem this is one that goes through multiple phases of the same substance this case we start with ice we have 50 grams ice I'm going to heat it from -10 degrees Celsius to steam at 122 1st I want to do what you going to want to try heating curve Chalmette start at -10 how to end up at a hundred twenty degrees so when you're here. It's going to melt at 0 I just completely melted it's going to heat up to a hundred going to boil and then scan this seems going to heat up to 120 degrees shop we are working has five segments with his five equations are going to use each segment that's the temperature change when we label them up solid liquid and gas they're going to use the equation Q equals MC delta T the reason I do this is a temperature change equations for heated up temperature changes now the plateau is where the phase changes occur and the bottom one is Fusion or melting and the top one is vaporization search equations for those cute girls mhv and cute mhf the reason is because during a phase change the temperature of the change so if you put a zero in there with just screw everything up now you have all five equations to plug in the numbers start with m which means the mass can be added to each equation right now El granero Constance the specific heat this is seeing this isn't what right now this is a Christian because when the gas phase solo * 2.09 and that's joules per gram degree Celsius now I got to go to the next one HD HD is the heat of vaporization which is 2260 joules per gram and we're going to go to liquid water what are the specific heat is 4.18 joules per gram degrees Celsius and then we're going to go to the heat of fusion and heat of fusion is 334 joules per gram and the specific heat of the solid which is 2.09 joules per gram degree celsius now the heat of fusion and heat of vaporization calculations are done we have to figure out the Delta cheese for now this isn't hard because you can at your heating curve you just threw the gas only exist from 100 to 120 which means the temperature changes like 20 degrees Celsius for liquid water goes from zero to a hundred so it's a hundred degrees Celsius play icicle from -10 0 which is a 10 degrees Celsius change chownow the way of all five equations complete is the fun part we get to actually calculate so 50 * 2.09 20 that's 2090 joules 50 * 2260 that's 11 or 12 113000 joules 50 * 4.18 * 100 gives you 20900 joules and 50 * 334 at 16700 Jules and 50 * 2.09 * 10 that gives you 1045 Charleston at them up 20900 and 113 in 1000 + 2090 + 850 3735 Jules or monthly 154 kilojoules of energy is required to take 50 grams of ice at negative and turn to steam at 120 degrees Celsius there's other ways of doing this you can actually use like distribution everything but this might be a little longer but it gets the job done. You can see what you're doing and that's complex Colorado

Uploaded by Debra Hood.

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Solving Complex Calorimetry Problems involving changes of state and temperature

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