Question

A metal block of density 6000 kg $m^{-3}$ and mass 1.2 kg is suspended through a spring of spring constant 200 N $m^{-1}$. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260 g and the block is at a height 40 cm 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 250 J $k{g}^{-1}{K}^{-1}$ and that of water is 4200 J $k{g}^{-1}{K}^{-1}$. The heat capacities of the vessel and the spring are negligible.

Answer 1

Volume of the block = $\frac{1.2}{6000}=2\times {10}^{-4}{m}^3$

When the mass is diped in water the block experinces a buoyant force and spring experience PE which is counteracted by its own weight.

$kx+V\rho g=mg$

$\Rightarrow 200x+2\times {10}^{-4}\times 1000\times 10=12$

$\Rightarrow x=\frac{(12-2)}{200}$

= $\frac{10}{200}$= 0.05

Now the heat is equally transfered to both block and water

So, $\frac{1}{2}k{x}^2+mgh-V\rho gh=m_1{s}_1\Delta \theta +m_2{s}_2\Delta \theta$

$\Rightarrow \frac{1}{2}\times 200\times 0.0025+1.2\times 10\times \left(\frac{40}{100}\right)$

$-2\times {10}^{-4}\times 1000\times 10\times \left(\frac{40}{100}\right)$

= $\left(\frac{260}{1000}\right)\times 4200\times \Delta t+1.2\times 250\times \Delta t$

$\Rightarrow 0.25+4.8-0.8=1092\Delta t+300\Delta t$

$\Rightarrow 1392\Delta t=4.25$

$\Rightarrow \Delta t=\frac{4.25}{1392}$ = 0.0030531

= $3\times {10}^{-30}$ C.