A thermos with a capacity of V = 3L was filled with boiling water and closed.

A thermos with a capacity of V = 3L was filled with boiling water and closed. A day later, the water temperature became t (k) = 80’C. How much warmth has the water given during this time?

In order to calculate the amount of heat given by the water per day, after cooling from the initial temperature of 100 degrees to a temperature of 80 degrees, we will use the formula: Q = m • c • Δt. The mass of water can be found by the formula: m = ρV, where the density of water is ρ = 1000 kg / cubic meter, and the volume is V = 3 liters = 3 cubic dm = 0.003 cubic meters. We get the mass m = 3 kg (1 kg of water takes up a volume of 1 liter). We find the specific heat capacity of water in the table of specific heat capacities of substances in physics reference books or on the Internet: c = 4200 J / (kg • deg), temperature change Δt = 100 – 80; Δt = 20 degrees. Substituting the values ​​of physical quantities in the formula, we get Q = 3 kg • 4200 J / (kg • deg) • 20 deg = 252 000 J = 252 kJ.
Answer: the water in the thermos gave off 252 kJ of heat per day.