# 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.