Asked by katie
If 74.5 grams of a hot metal was added to 56.4 grams of water and the temperature of the water rose from 20.4 to 33.4 degrees, how many KJ of energy was lost by the hot metal?
Answers
Answered by
DrBob222
q = mass water x specific heat water x (Tfinal-Tinitial)
Answered by
katie
i need help setting this problem up. i don't understand which numbers go where
Answered by
DrBob222
There is no setting up. Just substitute the numbers into the equation.
q = your answer.
mass water = 56.4g in the problem.
specific heat water = look this up in your text or notes. It is 4.184 J/gram I think.
Tfinal = 33.4 from the problem.
Tinitial = 20.4 from the problem.
q will be in joules, the problem asks for kJ; therefore, divide the answer you get by 1000 to convert J to kJ.
q = your answer.
mass water = 56.4g in the problem.
specific heat water = look this up in your text or notes. It is 4.184 J/gram I think.
Tfinal = 33.4 from the problem.
Tinitial = 20.4 from the problem.
q will be in joules, the problem asks for kJ; therefore, divide the answer you get by 1000 to convert J to kJ.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.