Asked by Nicole
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.
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.
Answers
Answered by
Anonymous
Q=mcΔT
Q=mc(Tf-Ti)
(Tf-Ti)=Q/mc
Tf=(Q/mc)+Ti
Tf=(479/(89.0)(4.184))+8.00
Q=mc(Tf-Ti)
(Tf-Ti)=Q/mc
Tf=(Q/mc)+Ti
Tf=(479/(89.0)(4.184))+8.00
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