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QPage 47: Q18: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q18 Solution: A metal block of density and mass 1.2 kg is suspended through a …

A metal block of density 600 kg m−3600 \text{ kg m}^{-3} and mass 1.2 kg is suspended through a spring of spring constant 200 N m−1200 \text{ 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 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−1 K−1250 \text{ J kg}^{-1} \text{ K}^{-1} and that of water is 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1}. Heat capacities of the vessel and the spring are negligible.

Step-by-Step Verified Solution

**Answer:** Rise in temperature = 1.8×10−41.8 \times 10^{-4} �C. **Step-by-Step Explanation:** Given: Density of metal block = 600 kg/m_; Mass = 1.2 kg; Spring constant = 200 N/m; Mass of water = 260 g; Height = 40 cm; Specific heat capacity of block = 250 J/kg-K; Specific heat of water = 4200 J/kg-K. Volume of block = 1.2/600=2×10−31.2/600 = 2 \times 10^{-3} m_. When support breaks, the block falls and the spring potential energy is converted into thermal energy. Potential energy lost = mgh+12kx2mgh + \frac{1}{2}kx^2 (where xx is the extension of spring). Thermal energy = 1.2×10×0.4+12×200×(0.4)2=4.8+16=20.81.2 \times 10 \times 0.4 + \frac{1}{2} \times 200 \times (0.4)^2 = 4.8 + 16 = 20.8 J. Rise in temperature: ΔT=20.8(1.2×250)+(0.26×4200)=20.8300+1092=0.0149�C\Delta T = \frac{20.8}{(1.2 \times 250) + (0.26 \times 4200)} = \frac{20.8}{300 + 1092} = 0.0149�\text{C}.

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