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?
3 answers
q = mass water x specific heat water x (Tfinal-Tinitial)
i need help setting this problem up. i don't understand which numbers go where
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.
0 answers