Water falls from a height 500 m. What is the raise in temperature of water at bottom. If the whole energy remains in the water? (Specific heat of water 4200 J / kg ° C ):

Question

Water falls from a height 500 m. What is the raise in temperature of water at bottom. If the whole energy remains in the water? (Specific heat of water 4200 J / kg ° C ): (A)  0.19° C (B)  2.3° C (C)  0.59° C (D)  1.17° C

Tardigrade Admin · Accepted Answer

Here: h=500 mg =9.8 m / s 2 c=4200 J / kg ° C Using the relation m g h=m c Δ θ m × 9.8 × 500=m × 4200 × Δ θ So, Δ θ=(9.8 × 500/4200) =1.17° C

Q. Water falls from a height $500 m$ . What is the raise in temperature of water at bottom. If the whole energy remains in the water?
(Specific heat of water $4200 J / k g^{\circ} C$ ):

$0.1 9^{\circ} C$

$2. 3^{\circ} C$

$0.5 9^{\circ} C$

$1.1 7^{\circ} C$

Here: $h = 500 m g = 9.8 m / s^{2}$
$c = 4200 J / k g^{\circ} C$
Using the relation $m g h = m c Δ θ$
$m \times 9.8 \times 500 = m \times 4200 \times Δ θ$
So, $Δ θ = \frac{9.8 \times 500}{4200}$
$= 1.1 7^{\circ} C$

Q. Water falls from a height 500m. What is the raise in temperature of water at bottom. If the whole energy remains in the water? (Specific heat of water 4200J/kg∘C ):

0.19∘C

2.3∘C

0.59∘C

1.17∘C