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.

Question

Infinity Learn · Accepted Answer

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

$\begin{array}{l} 200 x + \frac{2}{5000} \times 1000 \times 10 = 2 \times 10 \\ or x = \frac{20 - 4}{200} = \frac{16}{200} = 0 .08 m = 8 cm \end{array}$

Thus in equilibrium, energy stored in spring is

$U = \frac{1}{2} {kx}^{2} = \frac{1}{2} \times 200 \times (0 .08)^{2} = 0 .64 Joule$

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

$\begin{array}{l} 0 .64 + mg (0 .4) = m_{ω} \times s_{ω} \times ΔT + m_{B} \times s_{B} \times ΔT \\ or 0 .64 + 2 \times 10 \times 0 .4 = [0 .3 \times 4200 + 2 \times 234] ΔT \\ or ΔT = \frac{8 .64}{1768} = 0 {.005}^{\circ} C \end{array}$

Calorimetry

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material