A metal block of density 5000 kg/m3 and mass 2 kg is suspended by a spring of force constant 200 N/m. The spring block system is submerged in a water vessel. Total mass of water in it is 300 g and in equilibrium the block is at a height 40 cm above the bottom of vessel. If the support is broken. Find the rise in temperature of water (in oC). Specific heat of the initial of block is 250 J/kg K and that of water is 4200 J/kg K. Neglect the heat capacities of the vessel and the spring.

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 i.e. weight of block minus the buoyant force on block. Thus for equilibrium of block in water, we have kx + weight of liquid displaced = weight of block     200x+25000×1000×10=2×10or  x=20−4200=16200=0.08m=8cmThus in equilibrium, energy stored in spring isU=12kx2=12×200×(0.08)2=0.64 Joule When the support is broken, the mass falls down to the bottom of vessel 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      0.64+mg(0.4)=mω×sω×ΔT+mB×sB×ΔTor  0.64+2×10×0.4=[0.3×4200+2×234]ΔTor  ΔT=8.641768=0.005∘C