If 60 % of the kinetic energy of water fallin | vedclass

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(B) Given: Height $h = 210 \ m$,Specific heat $c = 4.2 	imes 10^3 \ J \ kg^{-1} \ K^{-1}$,Acceleration due to gravity $g = 10 \ m/s^2$.When water fal

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$0.3^{\circ} C$

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(B) Given: Height $h = 210 \ m$,Specific heat $c = 4.2 	imes 10^3 \ J \ kg^{-1} \ K^{-1}$,Acceleration due to gravity $g = 10 \ m/s^2$.When water falls,its potential energy is converted into kinetic energy.Potential Energy $(PE) = mgh$.According to the problem,$60 \%$ of the kinetic energy is converted into heat energy $(Q)$.$Q = 0.6 	imes PE = 0.6 	imes mgh$.Also,the heat produced is given by $Q = mc\Delta T$,where $\Delta T$ is the rise in temperature.Equating the two expressions for $Q$:$mc\Delta T = 0.6 	imes mgh$.Canceling mass $m$ from both sides:$c\Delta T = 0.6 	imes g 	imes h$.$\Delta T = \frac{0.6 	imes g 	imes h}{c}$.Substituting the values:$\Delta T = \frac{0.6 	imes 10 	imes 210}{4.2 	imes 10^3} = \frac{1260}{4200} = 0.3^{\circ} C$.Thus,the rise in temperature is $0.3^{\circ} C$.

If $60 \%$ of the kinetic energy of water falling from a $210 \ m$ high waterfall is converted into heat,the rise in temperature of the water at the bottom of the fall is nearly (specific heat of water $= 4.2 \times 10^3 \ J \ kg^{-1} \ K^{-1}$ and $g = 10 \ m/s^2$):

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