A 67 g piece of aluminum at 725K is dropped into a 165 g of H2O...

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)

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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