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Thermal Physics

Hot goIITian

Joined:	22 Sep 2007
Post:	118

21 Jan 2008 23:48:38 IST

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HC Verma calorimetry

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18) A metal block of density 6000kg/m^3 and mass 1.2kg is suspended through a spring constant 200N/m.The spring-block system is dipped in water kept in a vessel. The water has a mass of 260g and the block is at a height 40cm above the bottom of the vessel. If the support to the spring is broken,what will be the rise in the temperature of the water.Specific heat capacity of the block is 250J/kg-K and that of water is 4200 j/kg-K.Heat capacities of the vessel and the spring are negligible.

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svj 29

Blazing goIITian

Joined: 27 Dec 2007

Posts: 822

22 Jan 2008 10:42:11 IST

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When the block is in equilibrium in the water and spring is stretched by a distance x and the spring force balances the effective weight of block..ie weight of block minus the buoyant force on block..thus for equilibruim of block in water, we have:
kx + weight of liquid displaced = weight of block

200x + 1.2/6000 into 1000 into 10 = 1.2 into 10

solve n u'll get x=1/20m=5cm

thus equilibruim enrgy stored in the spring is
U=1/2 k into x square...
=1/2 into 200 into (1/20) square...
=0.25J

when the support in being broken, the mass falls down to the bottom of the vesel and the potential energy stored in spring and the gravitational potential energy of block is released and used in heating the water and block. thus we have:

0.25 + mg(0.4) =mass of watr into specific heat capacity of water
into change in temp. + mass of block into specific
heat capcity of blok into change in temp...

0.25 + 1.2 into 10 into 0.4 = (4200 into 0.26 + 250 into 1.2)into dlta T

5.05= delta T(1092 + 300)

dlta T = 5.05/1392 = 0.003 degree C.......

hope dis hlpd ya........

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