A piece of titanium metal with a mass of 20.8 g is heated in boiling water to 99.5 0C and then dropped into a coffee cup calorimeter containing 75.0 g of water at 21.7 0C.When thermal equilibum is reached, the final temperature is 14.30C.Calculate the specific heat capacity of titanium. ( Specific Heat Capacity of H2O (l) =4.184 J g-1 0C-1)
Lets see the formula at thermal equilibrium
Mass of titanium specific heat capacity (change in temperature) + mass of water specific heat capacity of water (change in temperature) = 0
20.8 g specific heat capacity (14.30 C -99.50 C) + 75.0 g 4.183 J/g.0 C (14.30 C -21.70 C) = 0
-1772.16 specific heat capacity -2321.56= 0
specific heat capacity = (-2321.56) / (-1772.16)
= 1.31 J g-1.0C-1
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