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

Given:-

Density of metal block, d = 600 kg m⁻³

Mass of metal block, m = 1.2 kg

Spring constant of the spring, k = 200 N m⁻¹

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

When the mass is dipped in water, it experiences a buoyant force and in the spring there is potential energy stored in it.

If the net force on the block is zero before breaking of the support of the spring, then

kx + Vρg = mg

200x + (2 × 10⁻⁴)× (1000) × (10) = 12

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

${\displaystyle \Rightarrow x=\frac{10}{200}=0.05\text{m}}$

The mechanical energy of the block is transferred to both block and water. Let the rise in temperature of the block and the water be ΔT.

Applying conservation of energy, we get

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

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

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

${\displaystyle \Rightarrow 0.25+4.8-0.8=1092\Delta T+300\Delta T}$

${\displaystyle \Rightarrow 1392\Delta T=4.25}$

${\displaystyle \Rightarrow \Delta T=\frac{4.25}{1392}=0.0030531}$

${\displaystyle \Delta T=3\times {10}^{-3}°C}$