Water falls from a height of 100 m. Find the rise in temperature of the water when it reaches the bottom, assuming all its energy is converted to heat at the bottom of the waterfall. $(g = 10 m s^{- 2}, s_{w} = 4200 J k g^{- 1} K^{- 1})$

Open in App

Solution

Given:
Specific Heat of water, $s_{w} = 4200 J / k g K$
$g = 10 m s^{- 2}$ , the acceleration of gravity
Height, $H = 100 m$
According to Energy conservation,
$s_{w} Δ T = g \times H$
So, rise in temperature, $Δ T = \frac{s_{w}}{g H} = \frac{4200}{10 \times 100} = 4.2 K$

Suggest Corrections

Similar questions

Water falls from a height of 100 m. Calculate the rise in temperature of water when it strikes the bottom. Given, S $_{w}$ = 4200 Jkg $^{- 1}$ $^{0}$ C $^{- 1}$ and g = 10 m/s $^{2}$ .

Q. Water falls from a height of 50 cm. Calculate the rise in temperature of water when it strikes the bottom. (

g = 10 m s^{- 2}

, sp. heat capacity of water

= 4200 J / k g^{o} C

)

Water falls from a height of 50 m. Calculate the rise in temperature of water when it strikes the bottom. Take g = 10 m s^-2 and specific heat capacity of water = 4200 J kg^-1 °C^-1).

Q. Water falls from a height of 500 m. What is approximately the rise temperature of water at bottom if whole energy is used to heat the water?

Q. Water falls from a height of

210 m

. Assuming whole of energy due to fall is converted into heat the rise in temperature of water would be
(

J = 4.3

Joule/cal)

Join BYJU'S Learning Program

Water falls from a height of 100 m. Find the rise in temperature of the water when it reaches the bottom, assuming all its energy is converted to heat at the bottom of the waterfall. (g=10ms−2,sw=4200Jkg−1K−1)

Given:Specific Heat of water, sw=4200J/kgK g=10ms−2, the acceleration of gravity Height, H=100m According to Energy conservation,swΔT=g×HSo, rise in temperature, ΔT=swgH=420010×100=4.2K

Water falls from a height of 100 m. Find the rise in temperature of the water when it reaches the bottom, assuming all its energy is converted to heat at the bottom of the waterfall. $(g = 10 m s^{- 2}, s_{w} = 4200 J k g^{- 1} K^{- 1})$

Given:
Specific Heat of water, $s_{w} = 4200 J / k g K$
$g = 10 m s^{- 2}$ , the acceleration of gravity
Height, $H = 100 m$
According to Energy conservation,
$s_{w} Δ T = g \times H$
So, rise in temperature, $Δ T = \frac{s_{w}}{g H} = \frac{4200}{10 \times 100} = 4.2 K$