Respuesta :
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 = [tex]m \times C \times (T_{2} - T_{1})[/tex]
q = [tex]1500.0 g \times 4.184 J/g^{o}C \times (28.20 - 25.40)^{o}C[/tex]
= [tex]1.757 \times 10^{4} J[/tex]
Therefore, the aluminum lost will be [tex]-1.757 \times 10^{4} J[/tex]. Hence, for the aluminum:
q = [tex]m \times c \times (T_{2} - T_{1})[/tex]
[tex]-1.757 \times 10^{4} J = 73.5 g \times C \times (28.20 - 98.7)^{o}C[/tex]
C = [tex]\frac{-1.757 \times 10^{4} J}{73.5 g \times (-70.5^{o}C)}[/tex]
= [tex]0.0003390 \times 10^{4} J/g^{o}C[/tex]
= 3.39 [tex]J/g^{o}C[/tex]
Therefore, we can conclude that specific heat of aluminium under given conditions is 3.39 [tex]J/g^{o}C[/tex].
