An 80.0g sample of water at 10.0 degrees C absorbs 1680J of heat energy. What is the final temp of the water?

Question

(i cant figure out what i did wrong on this one.)

A)50.0 degrees C
B)15.0 degrees C
C)5.00 degrees C
D)4.00 degrees C

heres what i did:
q=mC(delta)T
1680=80(4.18)x
1680=334.4x
x=5.02...
so i put C, but that was wrong

please show me what to do for this
thanks

Answer 1

Q=mc(t2 - t1)
1680J = .08kg(4184J/KgC)(t2-10)
1680/[(.08)(4184)] = t2 - 10

5.089 = t2 - 10
15.089 = t2

Your answer is B.

T2 is your final temperature, and T1 is your initial temperature. You solved for the temperature change only, which is technically okay...as long as you remember to add the delta T to your initial temperature to find the final temperature. I write is as T2-T1 and expand it out etc...to avoid making that mistake. :-D

Answer 2

Looks like you did this right but missed one final step -- the delta.

if we replace the delta in the original formula we get:
q = m * C * (Tfinal - Tinitial)

then from your results above:
x = (Tfinal - Tinitial) = 5.02

We know the initial temperature is 10 degrees... so...
Tfinal - 10 = 5.02
Tfinal = 15 degrees C

As a final check, it makes sense that the temperature of the water has risen as we have added heat/energy to the water.