A slice of pizza has 500 kcal. If we could burn the pizza and use...

Question

A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what would be the approximate increase in the temperature of the water? (Note: A liter of cold water weighs about 1 kg.)
a. 50°C
b. 5°C
c. 100°C
d. 10°C

Accepted Answer

Hello everyone. So in today's video, we're going to be using the energy equation to calculate how much energy is required to increase the temperature of a sample of water by a certain amount of degrees. There's a very simple calculations question. So let's jump straight into it. We know that we need to use the energy equation because in a problem we are being asked how much energy is required to increase the temperature of a water sample by a certain amount, which is precisely what the hate energy question does. Now I want you to recall what this energy equation is and it's Q or the heat energy equals the mass times the specific times the change in temperature. Now, in our question prompt were only given that we have a six liter sample of water and we need to raise the temperature by five degrees. So we're given the change in temperature and the mass which is six kg as per this note where we have that one liter of water equals one one kg of water. Were being asked by the Heat Energy, but we're also not given the specific heat of this water sample. So let's try to use the first prompt to try to obtain the specific heat of the water sample that we can use it in the first prompt were given that burning a piece of coal produces 400 kilocalories of energy, which is the heat energy of the reaction. So let's write that down. We're also given that these reactions change the temperature of a 40 liter sample of water, which through these note down here, we can get the mass, which in this case is going to be 40 kg since one kg of water equals one liter of water. Finally, we are given that this process raises the temperature of the sample by 10 degrees. So we also have the change in temperature, which in this case is 10. So let's use all of these to calculate the specific heat and later use it to solve the question. So we have that 400 kg calories equals 40 kg times the specific heat, which we don't know times the change in temperature, which in this case is 10 degrees. If we proceed with this calculation, we're going together 400 kg calories equals 400 times the specific heat. And if we divide both sides by 400 we're going to see the sea or the specific heat equals one. So now that we have been able to determine the specific heat of water, let's use it in our actual question. So we already established that we have six liters of water which is equivalent to six kg of water or the mass of the sample. And we also established that the temperature change is five degrees and the specific heat is one. So let's plug all these into our new equation for Q. You have the Q equals the mass, which in this case is six kg tends one, which is the specific heat of water as we found Times the temperature change, which we are given that is 5°. If we perform this calculation, we can see that Q, or the hit required to perform this reaction is 30 and the unit is kilo calories, which looking our answer choices corresponds with answer choice C. So thank you very much for sticking around until the end of this video, and I hope to see you on the next one.

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