Q=mcΔT
Q=mc(Tf-Ti)
(Tf-Ti)=Q/mc
Tf=(Q/mc)+Ti
Tf=(479/(89.0)(4.184))+8.00
Energy in the amount of 479 J is added to a
89.0 g sample of water at a temperature of
8.00◦C. What will be the final temperature of
the water?
Answer in units of ◦C.
1 answer