posted by asra on .
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?
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