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

b) phosphorus pentoxide