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

i ahve no idea what the first two answerers meant. could someone do out the math for me? that might help.

i still dont get it!!

Answer 1

You solved for the "delta-T" -- the change in temperature. You started a 10 degrees, and you gained 5, giving you a final temp of 15 degrees. Remember, they're not asking for the temperature change, they want the final temperature.

Answer 2

What their trying to say is you misconceived the variable Delta T. In other words you thought the value you solved for was the temperature when in fact you solved for the change in temperature. You did your math right, but you didn't finish your work. Take the value for delta t (5) and add it to the initial temperature (10) and you will have the final temperature of B) 15.00


Delta T is one variable. It's the change in temperature. (5 degrees C)
C is the specific heat capacity. (4.18)
m is the mass of the substance. (80)
q is the heat. (1680J)

Answer 3

The water started at 10.0degrees C, so you need to add the 5degrees C to initial temp.