specific heat capacity solving for joules example how much energy must be absorbed by 20 grams of water to increase its temperature from 283 degrees Celsius to 303 degrees Celsius the specific heat capacity of water is 4.18 joules per gram degrees Celsius so the first thing we're going to do is write our formula Q equals MC delta T so I look through my information here and we're solving for Q because they asked for how much energy so we're going to solve for Q the mass here is 20 grams of water we find that the specific heat capacity of water is listed so that's our C 4.18 joules per gram degrees Celsius and then they give us two temperatures and when we look at these temperatures here they give us a temperature initial and they give us a temperature final now remember to find delta T I'm going to take temperature final minus temperature initial so if I write down the final 10 temperature 303 and I subtract that from 283 we find the difference to be 20 degrees Celsius so for my delta T over here I'm going to put in 20 degrees Celsius now just to make sure that we're on the right track here we know that grams will cancel grams degrees Celsius which is on the bottom here will cancel degrees Celsius up here and I'm left with joules as my final unit for my overall answer so I'm going to take Q is equal to 20 times four point one eight times 20 and my final answer here will be 1672 joules and if I was doing it to the correct number of significant figures which really should be three this should be 1600 in 70 joules as my final answer using three significant figures