To what temperature will 8200 J of heat raise 3.0 kg of water that is initially at $10.0 ^\circ \textrm{C}$?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The heat gained by a substance is its mass times the specific heat capacity times the final temperature minus the initial temperature. And we can distribute the mc into this bracket here and we get mc times T final minus mc times T initial. Bring this to the other side making it positive and then switch the sides around, you have mc times final temperature equals Q plus mc times initial temperature. Divide both sides by mc and you get final temperature is Q plus mc Ti over mc. So, Q is the heat gain, that's 8,200 joules. And mass is 3 kilograms, and the specific heat capacity of water is 4,186 joules per kilogram kelvin degree, times the initial temperature in absolute kelvin. So, 10 degrees Celsius plus 273 divided by the mass, 3, times the specific heat capacity. And that gives about 283.65 kelvin. And we subtract away 273 to get the answer in Celsius, 10.7 degrees Celsius.