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 =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
To find the specific heat of the metal, you can use the equation:
mCΔT(water) = mCΔT(metal)
Where:
m = mass of the metal
C = specific heat of the metal
ΔT(water) = change in temperature of the water
ΔT(metal) = change in temperature of the metal
In this case, we have:
m = 150g
ΔT(water) = final temperature of the system - initial temperature of the water = 23°C - 20°C = 3°C
ΔT(metal) = final temperature of the system - initial temperature of the metal = 23°C - 80°C = -57°C (Note: The change in temperature of the metal is negative because the initial temperature of the metal is higher than the final temperature.)
Substituting the values into the equation, we get:
150g * C * 3°C = 150g * C * -57°C
Next, we can solve for C:
3°C = -57°C
Divide both sides by 150g:
3°C / 150g = -57°C / 150g
C = -0.16 J/g°C
So, the specific heat of the metal is -0.16 J/g°C.
Note: The negative sign indicates that the metal releases heat to the water as it cools down.