 Okay, so this is a pretty steppy or complex problem. Heating 56 liters of water from 85 degrees Fahrenheit to 212 degrees Fahrenheit, that's going to take a lot of energy. But for a chemistry problem, it's going to be pretty complex because Fahrenheit is not the units you want temperature to be in, you want those in Celsius. 212 degrees Fahrenheit, if you don't know, that's the boiling point of water in degrees Fahrenheit. So that one should be pretty easy to convert over to 100 degrees Celsius, which you should know is the boiling point of water in degrees Celsius. But I do have the formula there written for you for the TF. But for TI you do have to calculate it. So TI equals 5 9 times 85 minus 32 there, and that's going to give you 29 degrees Celsius. And like we said before, 100 degrees Celsius is the boiling point of water. I have that little dot there after the hundred to hundred the number because both of those zeros are significant and we have to indicate that. Okay, so next we do the change in temperature of the water. So that's the final temperature minus the initial temperature. And we see that that's 71 degrees, 100 degrees Celsius minus 29 degrees Celsius is 71 degrees Celsius. And then next we have to get the mass of water. So in order to get the mass of water we have to use that volume of water that we were given, 56 liters. And we have to remember that for every one liter there's a hundred mils, so cancel out our liters. And then from there cancel out our mils by remembering the density of water is one gram per mil. So we're going to get that number 56,000 grams. I'm going to put it into scientific notation just to make it easier on us. You could use either one of those numbers. But scientific notation makes it less cumbersome in my opinion. Okay, so remember the specific heat of water that's given to you, 4.184 joules per grams degrees Celsius. So those units are important. So you can cancel out the rest of your units. And we're looking for how much heat is needed to heat up that water. So in order to do this you're going to use the specific heat equation. So q equals mc delta t. And we've gotten all we've solved for all of those variables already. Just look back up and then just plug those numbers into where they go in the formula. So the mass is 5.6 times 10 to the fourth grams, the specific heat of water, and then the change of temperature. And then as you can see the grams and degrees Celsius cancel out leaving us with joules. And joules is a great unit of energy especially because we want to get to kill the joules at the end of the problem. So our answer is 1.7 times 10 to the 7th joules. And then the last thing we have to do is convert from joules to kilojoules. So remember for every 1000 joules there's one kilojoule. And so our final answer is 1.7 times 10 to the fourth kilojoules. So just keep in mind everything specifically making sure your units are correct for the problem and lastly your safe fix. You've got to remember your formulas for this one too. Let me know if there's any questions on that.