Here I calculate the energy needed to cool a beverage. I also estimate the cost of this energy.

I start by calculating the sensible heat that must be extracted from a can or bottle.

```
volume_l = 0.5
temperature_difference_K = 25.0
density_water_g_per_l = 1000.
heat_capacity_water_J_gK = 4.1855
kWh_per_joule = 1 / 3600000.
sensible_heat = 1
sensible_heat *= volume_l
sensible_heat *= temperature_difference_K
sensible_heat *= density_water_g_per_l
sensible_heat *= heat_capacity_water_J_gK
print('sensible heat (joules) =', sensible_heat)
print('sensible heat (kWh) = %0.3f' % (sensible_heat * kWh_per_joule))
```

The energy we must supply is less than this because typical refrigerators can remove more units of heat energy than the units of electrical energy delivered. Even with very expensive electricity ($1/kWh) this would add less than 0.01 USD to the cost of the drink.

This is the cost to cool down the drink. We can also estimate the cost to keep the drink stored at a cool temperature.

Assumptions:

- Refrigerator energy use as stated on product sheet

Each beverage has an average residence time in the refrigerator. We can estimate the energy use per can for storage by

$$ \frac{\textrm{energy used per unit time}}{\textrm{cans sold per unit time}} $$```
# haier refrigerator consumes 0.8 kWh per day
energy_per_day_kWh = 0.8
cans_per_day = 10
storage_energy_per_can = energy_per_day_kWh / cans_per_day
print('storage kWh per can', storage_energy_per_can)
```

Depending on the number of cans sold per day, the energy used by the refrigerator to keep the beverage cool more than the energy necessary to cool the beverage to drinking temperature.