An Al sphere of diameter 5.01cm at 100°C is placed on a 50g Cu ring of d=5cm at 0°C. Determine the mass of the sphere.The two are allowed to attain an equilibrium temperature and the sphere...

An Al sphere of diameter 5.01cm at 100°C is placed on a 50g Cu ring of d=5cm at 0°C. Determine the mass of the sphere.

The two are allowed to attain an equilibrium temperature and the sphere passes through the ring at that temp.

coefficient of linear expansion of Al = 23x10^-6 Cu = 17x10^-6 

specific heat capacity of Al= 0.215cal/g/°C  Cu = 0.0923cal/g/°C

Asked on by meghana17

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Let the equilibrium temperature be x deg.C

Here what happens is Cu absorbs heat from Al and get heats up. Due to that the diameter of Cu rings increases. Al losses temperature since Cu absorbs it. So the diameter of Al sphere decreases. When diameters of both rings are equal the sphere passes through ring.

But to this happen coefficient of expansion for Cu should be greater than Al. So in the question it should be modified as;

Coefficient of linear expansion of Cu = 23x10^-6

and Al = 17x10^-6 

At the equilibrium;

Diameter of Cu ring = d1(1+ `alpha*theta` )

                             = 5(1+23x10^-6*(x-0))

                             = 5(1+23x10^-6*x)

 

Diameter of Al sphere = d2 (1+`alpha*theta`  )

                                = 5.01(1+17x10^-6 *(100-x)

 

5(1+23x10^-6*x) = 5.01(1+17x10^-6 *(100-x)

 

Solving this will give you;

x= 92.5 deg.C

The heat removed my Al is absorbed by Cu.

Let mass of Al sphere is m.

Using Q=mc (theta)

Heat removed by Al (Q1)  = m*0.215*(100-92.5)

Heat absorbed by Cu (Q2) = 50*0.0923*(92.5-0)

 

At equilibrium Q1 = Q2

 m*0.215*(100-92.5)= 50*0.0923*(92.5-0)

Solving this equation will give you m= 264.74g

So the mass of the sphere is 264.74g

Sources:

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