A thermos with a capacity of V = 3L was filled with boiling water and closed. A day later, the water temperature

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? Q = m * T * C = 3 * 20 * 4200 J = 252 000 J = 252 kJ (4200 J) from where did you get it?

To find the heat given up by water in a thermos to the environment, we will use the formula: Q = Sv * m * (tboil – t) = Sv * ρw * V * (tboil – t).

Variables: V is the volume of water in the thermos (V = 3 l = 0.003 m3); t – temperature every other day (t = 80 ºС).

Constants: Sv – specific heat of water (Sv = 4200 J / (kg * K)); ρw is the density of water in a thermos (ρw = 1000 kg / m3); tboil – boiling point (tboil = 100 ºС).

Let’s perform the calculation: Q = Sv * ρw * V * (tboil – t) = 4200 * 1000 * 0.003 * (100 – 80) = 252 * 10 ^ 3 J = 252 kJ.

Answer: Over the past day, water has given off 252 kJ of heat.