Thread: Specific heat

How much heat (kJ) is needed to raise the temperature of 100.0 grams of water from 25.0°C to 50.0°C?

Q=m x C x change in temp

Q= ?
m= 100.0g
C=?
tf 25.0 C - Ti 50.0 C

trouble setting this up

about 12 KJ

I came up with q= 100.0 x 4.18 x 25.0
10450.
10.5 kj

Originally Posted by Louisana1

I came up with q= 100.0 x 4.18 x 25.0
10450.
10.5 kj

yea u are right, i should have put 4.18