Question

A metal block of density 600 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 bloc is at a height 40 cm above the bottom of the vessel. If the support of the spring is broken, what will be the rise in the temperature of the water. Specific heat capacity of the block is 250 J kg−3 K−1 and that of water is 4200 J kg−1 K−1. Heat capacities of the vessel and the spring are negligible.

Answer 1

Explanation:

$Given:\\ \mathrm{d}=600 \mathrm{kg} \mathrm{m}^{-3} \\ m=1.2 \mathrm{kg} \\ \mathrm{k}=200 \mathrm{N} \mathrm{m}-1 \\ V=\frac{1.2}{6000}=2 \times 10^{-4} \mathrm{m}^{3}$

V is Volume of the block,  

k is Spring constant of the spring,  

d is Density of metal block,

m is Mass of metal block  

It experiences a buoyant force when the mass is dipped in water, and in the spring there is potential energy stored therein.

If the net force on the block is zero before the spring support breaks,

$\begin{aligned}&k x+V \rho g=m g\\&200 x+(2 \times 10-4) \times(1000) \times(10)=12\\&x=\frac{12-2}{200}\\&x=\frac{10}{200}=0.05 \mathrm{m}\end{aligned}$

The block's mechanical energy is converted into both block and water. Let the concrete temperature rise and the water be equivalent to ΔT.

Applying energy conservation, we get

$\frac{1}{2} k x^{2}+m g h-V \rho g h=m_{1} s_{1} \Delta T+m_{2} s_{2} \Delta T$

$\frac{1}{2} \times 200 \times 0.0025+1.2 \times 10 \times\left(\frac{40}{100}\right)-2 \times 10^{-4} \times 1000 \times 10 \times\left(\frac{40}{100}\right)$

$0.25+4.8-0.8=1092 \Delta T+300 \Delta T$

$\begin{aligned}&1392 \Delta T=4.25\\&\Delta T=\frac{425}{1392}=0.0030531\\&\Delta T=3 \times 10^{-3 \cdot 0} C\end{aligned}$