Alright lets take a jump back in time!  Do you remember about your jr. year in high school?  You know when you had your first chemistry class?  Well if you don't we are about to jog your memory!

 

So back in your jr. year you most likely learned about specific heat which is the amount of energy that is needed to heat 1 gram or ml by 1 degree C.  With this information the units can be converted to lb or gallons or liters or really whatever you need.  Below is the equation for the energy needed.

 

q = (m)(C)(T₂ - T₁)

 

q = Amount of energy needed

m = Mass of what is being heated

C = The specific heat of m 

T₁ = Intial temperature

T_{2 = Desired temperature} 

(T₂ - T₁) = also modeled as delta T or the change in temperature

 

All of the above variables are known except C and q.  C can be solved for using if we knew q.  q is extremely hard to find without a full lab becuase it would involve a caloriameter.  Chances are as a homebrewer you don't have or do you need this expensive piece of equipment.  Therefore we will use the C values someone else has already found for us, no need to reinvent the wheel.  

 

Specific Heat values the Homebrewer may need

 

Methanol: 1.4 J/g/C^o

Glycerin: 2.43 J/g/C^o

Vegatable oil: 1.67 J/g/C^o

Water: 4.18 J/g/C^o

 

Methanol: .34 BTU/lb/F^o

Glycerin: .58 BTU/lb/F^o

Vegatable oil: .4 BTU/lb/F^o

Water: 1 BTU/lb/F^o

 

Now for the how to use it part.  Lets suppose bob is designing a processor.  Bob will do a 50 gallon batch in 20^o weather.

 

q = (50(8))(.4)(120)

 

The first set of parenthesis converts 50 gallons to 400 lb or oil becuase oil weighs 8 lb per gallon.  This conversion is necassary as the specific heat is BTU per lb not per gallon.  The second set of parenthesis includes the specific heat of vegatable oil which is .4.  The third set shows the temperature in which the oil needs to increase by. 

 

q = 19200 BTU

 

Therefore Bob will need 19200 BTU to hincrease the temperature of his 50 gallons of oil by 120 degrees.  Of course this figure is assuming that no heat is lost from the oil during the time period in which it is heated. Bob will be heating this oil in a water heater which is very well insulated so we will figure a 5% energy loss.

 

19200(1.05)

 

 Therefore Bob should consider a heat source that can supply 20160 BTU.  Bob can easily find this in a gas heater but what about electricity?  And what about the amount of time in which his oil's tempature is increased by that 120 degrees.  When bob converts 20160 BTU to watts he get about 6000 watts/hr.  This means that he needs a 6000 watt heating element if he wants his 120 degree increase to happen in an hour.  If he is content with 2 hours then he only needs a 3000 watt heater.

 

I hope that this little dive into chemistry has been helpful.