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Howdy,

First we need to know the amount of water you're using, and second, the amount of grain. For example, I'll use 100 gallons of water and 200 lbs. of grain. Water is 1 Btu/lb f and malted barley is about 0.38 Btu/lb f.

Water weight (8.3 lb/gal) x 100 gallons = 830 lbs.

Water 830 lbs + Grain 200 lbs = 1030 lbs

Cp of mash is ((1 Btu/lb°F x 830lbs=830) + (.38 Btu/lb°F x 200lbs=76)=906) / (1030 lbs) = .88 Btu/lb°F

1030 x .88 x 154 (mash temp) = 139,585.6 Btu

Spent mash:

200 x .38 Btu/lb°F x 170 (lauter temp) = 12,920 Btu

But, seeing you mentioned 10% ABV, are you wanting to know the Btu/lb°F of the fermented mash and then the left over lees? If so, here ya go:

Wash: water + 10% ethanol (using .6 Btu/lb f for ethanol at 75°F) + grain husks / weight = SH

((1 Btu/lb°F x 830lbs=830) + (.6 Btu/lb°F x 103lbs=61.8) + (.38 Btu/lb°F x 97lbs=36.86)=928.66) / (1030 lbs) = .9 Btu/lb°F

1030 x .9 x 75 (wash temp) = 69,525 Btu

Lees if the mash is used in the ferment and then left over (did not account for yeast or water uptake):

97 x .38 Btu/lb°F x 75 (lees temp) = 2,764.5 Btu

Please note this does not account for water uptake on the grain, but it should be close enough for ruff numbers.

I live in Bend if you'd like to chat. I'm a full time marketing person but I've been brewing for a long time and I'm looking to get into distilling.

Thanks,

Ross Wordhouse

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