Water fails from a height 500 m. What is the rise in temperature of water at bottom if whole energy remains in the water?

{0.96}^{o}

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{1.02}^{o}

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{1.17}^{o}

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{0.23}^{o}

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Solution

The correct option is C ${1.17}^{o}$ C
${1.17}^{\circ} C$

$E = m g h = 1 K g \times 9.81 m / s^{2} \times 500 m = 4905 J / l$

The specific heat of water $= 4180 J / (K g K)$

Rise in temperature, $= \frac{4905}{4180} = {1.17}^{\circ} C$

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Water fails from a height 500 m. What is the rise in temperature of water at bottom if whole energy remains in the water?

The correct option is C 1.17oC1.17∘CE=mgh=1Kg×9.81m/s2×500m=4905J/lThe specific heat of water =4180J/(KgK)Rise in temperature, =49054180=1.17∘C

The correct option is C ${1.17}^{o}$ C
${1.17}^{\circ} C$

$E = m g h = 1 K g \times 9.81 m / s^{2} \times 500 m = 4905 J / l$

The specific heat of water $= 4180 J / (K g K)$

Rise in temperature, $= \frac{4905}{4180} = {1.17}^{\circ} C$