In: Chemistry
A 67 g piece of aluminum at 725K is dropped into a 165 g of H2O (l) at 298K in an insulated container at 1 bar pressure. Calculate the temperature of the system once equilibrium has been reached. (Assume that Cp is independent on temperature)
mass of Al =67g=67g
mass of water=165g
Heat lost by Al at equilibrium=Heat gained by water
Q=m*Cp* T ,where, m=mass of substance
Cp=heat capacity at const P
T =change in temperature
Cp(Al)=0.900 J/g C
Cp=4.186 J/g C
67 g* 0.900 J/g C * (T-452 C)=165g*4.186 J/g C* (T-25 C) (T is the equilibrium temperature)
(T-452 C)/(T-25 C)=165g*4.186 J/g C/67 g* 0.900 J/g C
(T-452 C)/(T-25 C)=11.454
T-452 C=11.454(T-25C)
T-452 C=11.454 T-286.35
T-11.454 T=-286.35+452.0
10.454T=165.65
T=165.65/10.454=15.846C
T =15.846C