The simplest approach here is to neglect radiative losses (small at these low temperatures) and use Newton's Law of Cooling as a simple model:

$$\dot{Q_{loss}}=UA\left(T_w-T_{env}\right)$$

where:

• $T_w$ is the water temperature and $T_{env}$ is the ambient temperature
• $U$ is the overall heat transfer coefficient and $A$ the jug's surface area.

The estimation of $U$, depending on the jug wall's composition, can be found here.

The simplest approach here is to neglect radiative losses (small at these low temperatures) and use Newton's Law of Cooling as a simple model:

$$\dot{Q_{loss}}=UA\left(T_w-T_{env}\right)$$

where:

• $T_w$ is the water temperature and $T_{env}$ is the ambient temperature
• $U$ is the overall heat transfer coefficient and $A$ the jug's surface area.

The estimation of $U$, depending on the jug wall's composition, can be found here.

The jug is made of steel plates with air between them (maybe there is another insulating material between the plates—I could not find out exactly)

In order to compute an estimate of $U$ the precise material dimensions and characteristics would have to be known.