- #1
mikep
- 43
- 0
A 199 g lead ball at a temperature of 80.9°C is placed in a light calorimeter containing 178 g of water at 24.5°C. Find the equilibrium temperature of the system.
i used the equation [tex]Q = mc \Delta T[/tex]
-(199g)(4.186J/g°C)(T - 80.9°C) + (178g)(0.13J/g°C)(T - 24.5°C) = 0
T = T final = 82.5°C
can someone please tell mw what i did wrong?
can someone also confirm if the specific heat for lead is 0.13J/g°C?
i used the equation [tex]Q = mc \Delta T[/tex]
-(199g)(4.186J/g°C)(T - 80.9°C) + (178g)(0.13J/g°C)(T - 24.5°C) = 0
T = T final = 82.5°C
can someone please tell mw what i did wrong?
can someone also confirm if the specific heat for lead is 0.13J/g°C?