A typical nuclear reactor produces about 1.0 MW of power per day. What is the minimum rate of mass loss required to produce this much energy?

Video transcript

Hey everyone in this example, we have an average nuclear reactor generating 1.2 megawatts of electricity daily. How much mass at least is lost per day to generate this amount of energy. So we're going to begin with the unit were given 1.2 megawatts and we would understand that this amount of electricity is produced each day and each day we should understand that we have 24 hours. So our next step is to use the following conversion factor to go from megawatts two kilowatt hours, where we recall that we have for one megawatt 10 to the third power kilowatt hours. Now we're able to cancel those units of megawatts and we're going to continue our conversion on to go from kilowatt hours, two units of jewels for energy. So we're going to recall that for one kilowatt hour we have 3.60 times 10 to the six power jewels of energy. So we're going to be able to cancel out kilowatt hours. We can also cancel out ours. We're left with units of joules per day as our units for the energy produced each day. And this is going to give us a value of 1.368 times to the 11th power jewels per day of electricity produced. Now we can relate this to our mass energy equivalence formula which we recall is mass equal to our energy divided by the speed of light squared. So solving four M we would say that our energy in the numerator above is 1.368 times 10 to the 11th Power. Now we want to recognize that this prompt is asking us the mass that is lost per day. And so we want our final answer to be in units of grams per day. It can either be grams per day or kilograms per day. And so we should recognize that the conversion factor where one jewel is equal to one kg times meters squared divided by seconds squared. And we're going to use these units in place of our unit jewels. So we would have kilograms times meters squared, divided by seconds squared. And this is still divided by day or per day and just so that it's clear we'll make this bigger. Now in our denominator we're going to plug in our speed of light, which we should recall is 3.0 times 10 to the eighth Power in units of meters per second and this is squared in our denominator. So now we want to cancel our units and right now we have squared meters, we can cancel out, we can also cancel out r squared seconds and we're left with units of kilograms per day for our mass. So this is going to give us a final result equal to 1.152 times 10 to the negative six power kilograms per day. So this would be our final answer here to complete this example. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.