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A metal block of density 6000 kg/m*3 and mass 1.2 kg is suspended through a spring of spring constant 200 N/m . The spring block system is dipped in water kept in a vessel . The water has mass of 260 g and the block is at height 40 cm above the bottom of the vessel . If the support to the string is broken , what will be the rise in the temperature of the water. Specific heat capacity of the block is 250 J/kg/K and that of water is 4200 J/kg/K . Heat capacities of the spring and the vessel are negligible.[ taken from concepts of physics by Dr.H.C.Verma]
The answer is 0.003.
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1 solution
Heat lost = heat gain (1.2 g-(1.2/6000 1000) g)h=(m1s1+m2s2)ΔT 1 10 .4=(.26 4200+1.2*250)ΔT .ΔT=.0029°C=.003°C