The formula for calculating the heat-up time of a still given the power input and volume of liquid is as follows:

Heat-up time = (Volume of liquid x Specific heat of liquid x Temperature rise) / Power input

In this formula:

- Volume of liquid is the amount of liquid you are heating up, measured in liters (L)
- Specific heat of liquid is the amount of heat energy required to raise the temperature of the liquid by one degree Celsius (C), measured in joules per gram per degree Celsius (J/g·°C)
- Temperature rise is the difference between the starting temperature of the liquid and the desired final temperature, measured in degrees Celsius (°C)
- Power input is the amount of energy being supplied to the still, measured in watts (W)

By plugging in these values, you can calculate the heat-up time required to raise the temperature of the liquid in your still to the desired final temperature, given the power input being used.

## Calculator:

## Specific heat capacity of Ethanol and Water Mixture

Water-Ethanol Mixture | Specific Heat Capacity (J/g·°C) |
---|---|

100% Water | 4.18 |

90% Water, 10% Ethanol | 3.85 |

80% Water, 20% Ethanol | 3.52 |

70% Water, 30% Ethanol | 3.19 |

60% Water, 40% Ethanol | 2.86 |

50% Water, 50% Ethanol | 2.53 |

45% Water, 55% Ethanol | 2.36 |

40% Water, 60% Ethanol | 2.19 |

35% Water, 65% Ethanol | 2.03 |

30% Water, 70% Ethanol | 1.87 |

25% Water, 75% Ethanol | 1.71 |

20% Water, 80% Ethanol | 1.55 |

15% Water, 85% Ethanol | 1.39 |

10% Water, 90% Ethanol | 1.23 |

5% Water, 95% Ethanol | 1.07 |

### Example 1: 5 gallon (18.9 L) still filled with a 10% ABV (alcohol by volume) sugar wash:

here’s an example of how to calculate the heat up time for a 5 gallon (18.9 L) still filled with a 10% ABV (alcohol by volume) sugar wash:

Assuming the specific heat capacity of the sugar wash is approximately 4.18 J/g·°C (similar to that of water), and that the heat source provides a constant 1500 watts of power:

- Calculate the mass of the sugar wash in grams: 5 gallons x 3.785 L/gallon x 1000 g/L = 18925 g
- Calculate the mass of ethanol in the sugar wash: 10% ABV x 18925 g = 1892.5 g
- Calculate the specific heat capacity of the sugar wash: 0.1 x 4.18 J/g·°C + 0.9 x 4.18 J/g·°C = 4.18 J/g·°C
- Calculate the total heat capacity of the sugar wash: 18925 g x 4.18 J/g·°C = 79152.5 J/°C
- Calculate the energy required to raise the temperature of the sugar wash by 1°C: 79152.5 J/°C ÷ 100°C = 791.525 J/°C
- Calculate the time required to heat the sugar wash from 20°C to boiling (assuming boiling point of 78.37°C for a 10% ABV solution): Energy required to heat sugar wash to boiling = 18925 g x (78.37°C – 20°C) x 4.18 J/g·°C = 4471637.15 J Time required = Energy required ÷ Power input = 4471637.15 J ÷ 1500 W = 2981.09 seconds or approximately 49.68 minutes

Keep in mind that this calculation is an estimate and may vary based on factors such as the efficiency of the heat source and the insulation of the still.

## Example 1: of how to use the heat-up time formula to calculate the heat-up time of a 200 liter still with 6 kW elements:

Let’s say you want to heat up 200 liters of liquid from 20°C to 85°C using 6 kW elements.

First, you need to know the specific heat of the liquid you’re heating. Let’s assume you’re heating water, which has a specific heat of 4.18 J/g·°C.

Next, you can plug in the values into the heat-up time formula:

Heat-up time = (Volume of liquid x Specific heat of liquid x Temperature rise) / Power input

Heat-up time = (200 L x 4.18 J/g·°C x (85°C – 20°C)) / 6,000 W Heat-up time = 14.93 hours (rounded to two decimal places)

This means that it will take approximately 14.93 hours to heat up 200 liters of water from 20°C to 85°C using 6 kW elements.

It’s important to note that this calculation assumes no heat loss during the heating process. In reality, heat loss may occur due to factors such as insulation, ambient temperature, and the efficiency of the still’s heating elements. These factors may affect the actual heat-up time required.

Sure, here’s an example of how to use the heat-up time formula to calculate the heat-up time of a 200 liter still with 6 kW elements for a mixture of 50% water and 50% ethanol:

Let’s say you want to heat up 200 liters of a mixture that is 50% water and 50% ethanol from 20°C to 85°C using 6 kW elements.

First, you need to know the specific heat of the mixture you’re heating. The specific heat of the mixture will be a weighted average of the specific heats of water and ethanol, where the weights are the proportions of water and ethanol in the mixture. The specific heat of water is 4.18 J/g·°C and the specific heat of ethanol is 2.44 J/g·°C.

Specific heat of mixture = (0.5 x Specific heat of water) + (0.5 x Specific heat of ethanol) Specific heat of mixture = (0.5 x 4.18 J/g·°C) + (0.5 x 2.44 J/g·°C) Specific heat of mixture = 3.31 J/g·°C

Next, you can plug in the values into the heat-up time formula:

Heat-up time = (Volume of liquid x Specific heat of liquid x Temperature rise) / Power input

Heat-up time = (200 L x 3.31 J/g·°C x (85°C – 20°C)) / 6,000 W Heat-up time = 10.48 hours (rounded to two decimal places)

This means that it will take approximately 10.48 hours to heat up 200 liters of a mixture that is 50% water and 50% ethanol from 20°C to 85°C using 6 kW elements.

Again, it’s important to note that this calculation assumes no heat loss during the heating process, and actual heat-up time may vary based on factors such as insulation, ambient temperature, and the efficiency of the still’s heating elements.

Here’s a worked example for a 200L still running at 6kW:

- Calculate the mass of the liquid in the still. Assuming the liquid has a density of 1 kg/L, the mass is:mass = volume x density = 200 L x 1 kg/L = 200 kg
- Determine the specific heat capacity of the liquid. For the purposes of this example, let’s assume the liquid is a sugar wash consisting of 50% water and 50% ethanol. The specific heat capacity of this mixture varies depending on the temperature and the exact composition, but a reasonable estimate is around 3.5 J/g·°C.
- Calculate the energy required to heat the liquid from room temperature (20°C) to boiling point (approximately 78.37°C for a 10% ABV solution). The temperature change is:ΔT = boiling point – room temperature = 78.37°C – 20°C = 58.37°CThe energy required is:energy = mass x specific heat capacity x ΔT = 200 kg x 3.5 J/g·°C x 58.37°C = 40,690,000 J
- Calculate the time required to heat the liquid using the formula:time = energy / powerwhere power is the power input of the still in watts, which in this case is 6,000 W.time = 40,690,000 J / 6,000 W = 6,781.67 secondsThis is equivalent to approximately 1.89 hours or 1 hour and 53 minutes.

So in this example, it would take approximately 1 hour and 53 minutes to heat a 200L still containing a sugar wash with 50% water and 50% ethanol from room temperature to boiling point using a 6kW heating element.