Question

A piece of titanium metal with a mass of 20.8 g is heated in
boiling water to 99.5 ^{0}C and then dropped into a coffee
cup calorimeter containing 75.0 g of water at 21.7
^{0}C.When thermal equilibum is reached, the final
temperature is 14.3^{0}C.Calculate the specific heat
capacity of titanium. ( Specific Heat Capacity of H_{2}O
(l) =4.184 J g^{-1 0}C^{-1})

Answer #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
(14.3^{0} C -99.5^{0} C) + 75.0 g 4.183
J/g.^{0} C
(14.3^{0} C -21.7^{0} C) = 0

-1772.16 specific heat capacity -2321.56= 0

specific heat capacity = (-2321.56) / (-1772.16)

= **1.31 J
g ^{-1}.^{0}C^{-1}**

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