Calculate the heat required to raise the temperature of 50.0g of water from 20.0C to 30.0C. Rob Lederer: This type of problem, quite a few steps. So, always like to draw a picture, not really a picture. It's just something that kind of helps me to understand what's going on. So, we're taking some ice at minus 20 degree Celsius and we warm it all the way to 30 degree Celsius. Here is the thing. We're stopping at Zero degree Celsius for a phase change. So we are driving along, stopping for the phase change and then going up to 30. Now, there are three different types of calculations that must be added here to figure out how much energy that 50 grams of water of ice actually needs in order to get from here to here. Minus 20 to zero and mcdeltaT because it's the temperature change. At zero, phase change NH. From zero to 30 temperature change mcdeltaT. So, in the formula heat lose equals heat gained, where the environment again something in the environment and surroundings is losing heat to this system, which is the ice cube, which is gaining the heat, is gaining the heat in three ways. two kinetic ways, two temperature changes and one phase change and you might say, well, yeah well, chemguy look, why don't you just go from minus 20 to 30 because that's one big temperature change there and the phase change involved there because the heat capacity for water as the solid as opposed to water as a liquid from 0 to 30 is different. So, you go to do a different mcdeltaT. Here it is. The heat lose equals the mcdeltaT for the ice warming up to zero plus the phase change, that's the phase change, when the water turns from ice to liquid and this of course is the temperature change for the warming of the water from zero all the way to 30. So this is the water liquid, this is the phase change and this is the water as a solid going from minus 20 to 0. Plug in numbers. The way it looks is this. This term is actually here, this term is here and this term is here. But I have them all together to come up with our answer. Now what I plugged in was, for the mass here and in here, the 50 grams but not in terms of grams, but that should be a kg. I turn them all into kilograms because, I know that the big H, the molar heat is in the unit kilojoules and so therefore I want to keep my answer in kilojoules in the end. You will have to do that. You can actually turn it into joules per mole and turn it out to 6.01 kilojoules per mole into 6.01 times 10 to the 3 joules per mole. but look, on diploma exams or major exams finals that I have seen, the one thing that gets in and that's almost intentionally done to see if you are paying attention, is to interchange numbers that will arrive at joules for one answer and may be kilojoules for another and they can't be directly added together. So you got to be so careful and make sure everything is in same unit. So I'm turning all the masses into kilograms. So the heat capacities are written as for water, as ice 2.00 kilojoules per kilogram degree Celsius. Remember that's in the joules. Now the kilograms cancel and we are left with kilojoules. And then, I know this sounds kind of weird, but stay with me. All of the temperature changes that you ever write must be, temperature changes must be a positive number, don't make it a negative. If you want a negative 20 in here as a temperature change, your number is going to be negative and then you are going to add it to two others and you are going to get less than you should have in terms of total amount of energy at the end. All deltaTs are positive. Now, look I always like to plug in initial and a final temperature. This one here being of course, the final temperature in this calculation, this is initial, initial minus final here. But the point is in which way you plug it in as a long as you can get the difference between these two is positive. So, look what I did, I know it kind of looks kind of goofy, but I said zero minus negative 20, now that makes it a positive n