A water heater is operated by solar power. If the solar collector has an area of 6.39 m2, and the intensity delivered by sunlight is 550 W/m2, how long does it take to increase the temperature of 1.00 m3 of water from 19.4∘C to 72.1 ∘C?

62979.09 s or 17.49 h

Explanation:

From the question,

The power from the solar needed to increase the temperature of water is given as,

P = P'A Equation 1

Where P = power, P' = intensity of sunlight, A = Area of the solar collector.

Given: P' = 550 W/m², A = 6.39 m².

Substitute into equation 1

P = 550(6.39)

P = 3514.5 W.

The amount heat required to change the temperature  of water is given as,

Q = cm(t₂-t₁) Equation 2

Where Q = amount of heat, c = specific heat capacity of water, m = mass of water, t₂ = Final temperature of water, t₁ = initial temperature of water

Given: c = 4200 J/kg.°C, m = Density×volume = 1000×1 = 1000 kg, t₂ = 72.1 °C, t₁ = 19.4 °C

Substitute into equation 2

Q = 4200(1000)(72.1-19.4)

Q = 4200(1000)(52.7)

Q = 221340000 J.

But,

P = Q/t

Where t = time.

Make t the subject of the equation,

t = Q/p Equation 3

t = 221340000/3514.5

t = 62979.09 s

t = (62979.09/3600) hrs

t = 17.49 h

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

pls mark as brainliest.