If 765000 J of energy are added to 7.90 L of

water at 291 K, what will the final temperature
of the water be?
Answer in units of K.

2 answers

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute and solve for Tf.
I think you use the formula q=mc(delta)t
with q being the amount of energy added, m being the mass of water,c being the specific heat capacity of water which is 4.18J/g*degrees Celsius and then you would solve for delta t which is your change in temperature... hope this helps somewhat :)
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