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# An aluminium can of cylindrical shape contains 500 cm^3 of water. The area of the inner cross section of the can is 125 cm^2. All measurements refer to 10°C. Find the rise in the water level if the temperature increase to 80°C. The coefficient of linear expansion of aluminium= 23x10^-6/°C and average coefficient of volume expansion of water = 3.2x10^-4/°C.

28 Points
11 years ago

Dear Cathy,

Ans:- Let the co-efficient of linear exoansion of aluminium is=a and the coeff of vol expansion of water=y The initial area of the bottom=A0 and initial vol of water is V0

The final area is=At=A0(1+2at)        ( Coeff of lateral expansion= 2 coeff of linear expansion)

The final vol of water=V0(1+yt)

The initial length of water is= 500/125=4cm

the final length is= V0(1+yt)/A0(1+2at)

Putting the values we get , the final length=4.0765 cm

Hence the increase in the water level is=(4.0765-4)=.0765 cm(ans)

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Soumyajit Das IIT Kharagpur

Amrin
11 Points
3 years ago
∆T=change in temperature=80-10=70℃∆T=V-500/3.2×10^-5×500∆T=V-500/16×10^-270×16×10^-2=V-50011.2=V-500V=511.2=final volume of waterExpansion of aluminium:∆T=A-125/125×23×10^-670=A-125/2875×10^-670×2875×10^-6=A-1250.20125=A-125A=125.20125=final area of aluminium after expansionHeight of water at 10℃=500/125=4Height of water at 80℃=511.2/125.20125=4.083Rise in water level=4.083-4=0.083cm