Question

A 5.00-m-diameter garden pond is 0.500 m deep. Solar energy is incident on the pond at an average rate of 400 W/m2. If the water absorbs all the solar energy and does not exchange energy with its surroundings, how many hours will it take to warm from 16.0 ∘C to 19.0 ∘C?

Answer 1

Density of water = = 1000 kg/m³

Specific heat of water = C = 4186 J/(kg.^oC)

Diameter of the garden pond = D = 5 m

Depth of the garden pond = H = 0.5 m

Volume of water in the garden pond = V

V = D²H/4

Mass of water in the garden pond = M

M = V

M = D²H/4

Initial temperature of the water in the pond = T₁ = 16 ^oC

Final temperature of the water in the pond = T₂ = 19 ^oC

Heat needed to warm up the water = H₁

H₁ = MC(T₂ - T₁)

Surface area the solar energy is incident on = A = D²/4

Average rate of solar energy incidence on the pond = I = 400 W/m²

Time taken to heat up the pond = t

Heat gained by the pond from the solar energy = H₂

H₂ = IAt

The heat energy gained by the pond from the solar energy is equal to the energy gone in heating up the water.

H₁ = H₂

MC(T₂ - T₁) = IAt

(D²H/4)C(T₂ - T₁) = I(D²/4)t

HC(T₂ - T₁) = It

(1000)(0.5)(4186)(19 - 16) = (400)t

t = 15697.5 sec

Converting from seconds to hours,

t = 15697.5/3600 hours

t = 4.36 hours

Time taken to increase the temperature of the water in pond from 16 ^oC to 19 ^oC = 4.36 hours