Heat absorbed chem question?

Question

How much heat will be absorbed by 320g of water when its temperature is raised by 35 C? The specific heat for water is 4.18J/g x C.

I would be able to do this but im very confused as to why there is no original temperature to find the difference in temperature. help! >.<

Answer 1

You don&#039;t need the starting temperature ... only that the temperature goes up by 35oC.

So heat absorbed =  Spec. heat X mass X change in temp

= 4.18 X 320 X 35 = 46816 J

Answer 2

Yah, but let&#039;s assume 25 C STP for the initial. Ask about this, your professor?

q = smt

q = (4.18 J/g x C)(320g)(35-25 C)

q = 13. KJ  or  13. x 10^3 J  (answer was 13376 J) I assumed two sig figs based on 35 C remember to use decimal places!

Answer 3

Wow dude sorry I&#039;m with ChemGuy I have two accounts! Don&#039;t arrest me!

Okay, considering all the trouble you&#039;ve been through, you SAID by 35C not TO 35C.

t = (60-25C)

answer = 47. KJ or 47. x 10^3 J  answer being 46816 J