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If \(60 \%\) of the kinetic energy of water falling from \(210 \mathrm{~m}\) high water fall is converted into heat. The raise in temperature of water at the bottom of the falls is nearly (specific heat of water \(=4.2 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\))
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Verified Answer
The correct answer is:
\(0.3^{\circ} \mathrm{C}\)
Given, \(h=210 \mathrm{~m}\)
Specific heat of water, \(c=4.2 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)
When water falls on the surface of earth, then its potential energy is converted into kinetic energy.
\(\therefore\) Kinetic energy \(=\) Potential energy
\(\mathrm{KE}=m g h\)...(i)
According to question, Heat produced \(=60 \%\) of KE
\(\begin{aligned}
\Rightarrow \quad m c \Delta T & =\frac{60}{100} \times m g h \quad \text { [from Eq. (i)] } \\
\Delta T & =\frac{0.6 g h}{c}=\frac{0.6 \times 10 \times 210}{4.2 \times 10^3}=0.3^{\circ} \mathrm{C}
\end{aligned}\)
Specific heat of water, \(c=4.2 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\)
When water falls on the surface of earth, then its potential energy is converted into kinetic energy.
\(\therefore\) Kinetic energy \(=\) Potential energy
\(\mathrm{KE}=m g h\)...(i)
According to question, Heat produced \(=60 \%\) of KE
\(\begin{aligned}
\Rightarrow \quad m c \Delta T & =\frac{60}{100} \times m g h \quad \text { [from Eq. (i)] } \\
\Delta T & =\frac{0.6 g h}{c}=\frac{0.6 \times 10 \times 210}{4.2 \times 10^3}=0.3^{\circ} \mathrm{C}
\end{aligned}\)
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