Asked by asra
a 150g of a metal at 80degree C is placed in 100cm3 of pure water at 20degree C. the final temperature of the system (metal+water) is 23degree C. what is the specific heat of the metal?
i got the ans -0.16 by using formula mCdelta T(water)=mCdeltaT(metal) but i am not sure if the ans should be negative?
i got the ans -0.16 by using formula mCdelta T(water)=mCdeltaT(metal) but i am not sure if the ans should be negative?
Answers
Answered by
Elena
Q1 =Q2
Q1 = c(m)•m1•(t1 - t),
Q2 = c(w)•m2•(t - t2) = c(w)•ρ•V•(t - t2),
c(m) = c(w) •ρ•V• (t - t2)/ m1•(t1 - t) =
= 4180•1000•100•10^-6•(23-20)/0.150•(80-23) = 147 J/kg•K
Q1 = c(m)•m1•(t1 - t),
Q2 = c(w)•m2•(t - t2) = c(w)•ρ•V•(t - t2),
c(m) = c(w) •ρ•V• (t - t2)/ m1•(t1 - t) =
= 4180•1000•100•10^-6•(23-20)/0.150•(80-23) = 147 J/kg•K
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.