In a calorimet containing 1 liter of water, the propeller blades rotate due to the energy of two descending weights weighing

In a calorimet containing 1 liter of water, the propeller blades rotate due to the energy of two descending weights weighing 8.4 kg each. This causes the water to heat up. How many degrees did the water heat up when each weight dropped 10 m?

Given:
V = 1l = 0.001m ^ 3,
ρ = 1000kg / m ^ 3 – water density,
m = 8.4kg,
n = 2,
h = 10m;
Find: Δt -?
According to the law of conservation of energy, loads perform work due to the lost potential energy:
A = n * m * g * h,
where g = 10m / s ^ 2 is the acceleration of gravity;
If there are no losses, then all the work is converted into heat for heating water in the calorimeter:
A = Q = M * c * Δt,
where M = ρ * V is the mass of water,
c = 4200J / (kg * deg) – specific heat capacity of water;
From here we get:
Δt = A / (ρ * V * c) = n * m * g * h / (ρ * V * c);
Δt = 2 * 8.4 * 10 * 10 / (1000 * 0.001 * 4200) = 0.4 ° C.