How much heat does the water in the pond lose when cooled by 5 degrees? The pond area is 420 m, the depth is 1.4 m.

Initial data: Δt (change in water temperature in the pond) = 5 ºС; S (pond area) = 420 m ^ 2; h (pond depth) = 1.4 m.

Reference data: ρw (water density) = 1000 kg / m ^ 3; C (specific heat of water) = 4200 J / (kg * K).

1) The volume of water in the pond (we assume that the pond has the form of a rectangular parallelepiped): V = S * h = 420 * 1.4 = 588 m ^ 3.

2) The mass of water in the pond: m = ρw * V = 1000 * 588 = 588,000 kg.

3) Heat loss of water: Q = C * m * Δt = 4200 * 588000 * 5 = 12.348 * 10 ^ 9 J = 12.348 GJ.