If 60 % of the kinetic energy of water falling from 210 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 × J kg -1 K -1)

Question

If 60 % of the kinetic energy of water falling from 210 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 × J kg -1 K -1) (A) 0.6° C (B) 0.3° C (C) 1.2 K (D) 2.4 K

Tardigrade Admin · Accepted Answer

Given a , h=210 m Specific heat of water, c=4.2 × 103 J kg -1 K-1 When water falls on the surface of earth, then its potential energy is converted into kinetic energy. ∴ Kinetic energy = Potential energy K E=m g h ldots ldots.(i) According to question, Heat produced =60 % of K E [from Eq. (i)] ⇒ m c Δ T=(60/100) × m g h Δ T=(0.6 g h/c)=(0.6 × 10 × 210/4.2 × 103) =0.3° C

Q. If $60%$ of the kinetic energy of water falling from $210 m$ 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 $= 4.2 \times J k g^{- 1} K^{- 1}$ )

$0. 6^{\circ} C$

$0. 3^{\circ} C$

$1.2 K$

$2.4 K$

Q. If 60% of the kinetic energy of water falling from 210m 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×Jkg−1K−1)

0.6∘C

0.3∘C

1.2K

2.4K

Q. If $60%$ of the kinetic energy of water falling from $210 m$ 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 $= 4.2 \times J k g^{- 1} K^{- 1}$ )

$0. 6^{\circ} C$

$0. 3^{\circ} C$

$1.2 K$

$2.4 K$