Calculation of the natural cooling temperature of water in a container

The first thing to take into account is that this is a case of natural cooling, and this cooling process follows Newton’s law of cooling: ΔQ/Δt = h·A·(Twater - Tair), where ΔQ represents the heat transferred, Δt is the time, h is the heat transfer coefficient, A is the surface area, Twater is the temperature of the water, and Tair is the temperature of the air. This question asks us to find the water temperature 14 hours later, that is, to determine the change in water temperature during this cooling process. The factors we need to consider include the specific heat capacity C of water, the initial temperature Ti of water, the mass m of water, the heat transfer coefficient h, the surface area A, and the ambient temperature Tair. According to Newton’s law of cooling, we can obtain the following formula: dT/dt = -h·A·(T - Tair)/(m·C). By solving this differential equation, we get: T = Tair + (Ti - Tair) * exp(-h·A·t/(m·C)). In this problem, we know that: the specific heat capacity of water C = 4186 J/(kg·°C), the initial temperature of water Ti = 80°C, the mass of water m = 0.5 m³ * 1000 kg/m³ = 500 kg, the heat transfer coefficient h (the heat transfer coefficient for 304 stainless steel is approximately 20 W/(m²·°C)), the surface area A = 30 m², and the ambient temperature Tair = 15°C. t represents time; here it is 14 hours, which needs to be converted into seconds, so t = 14*3600 = 50400 seconds. By substituting these values into the formula, we can calculate the water temperature T after 14 hours. It should be noted that this formula is merely an idealized model; the actual situation may be influenced by other factors, such as the shape of the container and the convection of water. Therefore, this result can only serve as a rough reference. .