Question

73.5 g of aluminum is heated in boiling water to a temperature of 98.7C. the Aluminum is then placed in a calorimeter containing 1500.0 g of water at a temperature of 25.40C. The temperature of the water in the calorimeter increase to a final temperature of 28.20C. What is the specific heat of the Aluminum

Answer 1

Explanation:

Let us assume that if the calorimeter is ideal and absorbs no heat, then heat absorbed by the water will be as follows.

                          q = $m \times C \times (T_{2} - T_{1})$

                 q = $1500.0 g \times 4.184 J/g^{o}C \times (28.20 - 25.40)^{o}C$

                   = $1.757 \times 10^{4} J$

Therefore, the aluminum lost will be $-1.757 \times 10^{4} J$. Hence, for the aluminum:

                         q = $m \times c \times (T_{2} - T_{1})$

  $-1.757 \times 10^{4} J = 73.5 g \times C \times (28.20 - 98.7)^{o}C$

                         C = $\frac{-1.757 \times 10^{4} J}{73.5 g \times (-70.5^{o}C)}$

                            = $0.0003390 \times 10^{4} J/g^{o}C$

                            = 3.39 $J/g^{o}C$

Therefore, we can conclude that specific heat of aluminium under given conditions is 3.39 $J/g^{o}C$.