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.

Density of block = 6000 kg/m^3Mass = 1.2Kg => Volume = 0.0002 m^3Buoyant force on block(Fb) = (Volume of block)× (density of water)× g=> Fb = 0.0002×1000×10 = 2N (appx)Weight of block = mg = 12N (appx)=> Net force on spring = mg-(Fb) = 10N (appx)or kx = 10 => x = 10÷200 = 0.05 MTherefore, elastic potential energy of spring = 0.5kx^2 = 0.5 × 200 × 0.05^2 = 0.25JGravitational potential energy of block = mgh = 1.2 × 10 × 0.4 = 4.8J=> Total energy = 5.05JNow, Total energy = Heat of system = Internal energy of block + Internal energy of water => 1.2 × 250 × ∆T + 0.26 × 4200 × ∆T = 5.05JTherefore, ∆T = 5.05 ÷ (300+1092) = 0.003 ℃ or 0.003 K (appx)