Question

if 80% of the heat of heat from the combustion of a briquette with weight 93.18g makes it into a pot of boiling water, how many gallons of 72°f water can be boiled?

note water boils at 212°f and there are 8.3 pounds of water per gallon
biomass averages 8000 BTUs per pound.

Answer 1

Assume the biquette is essntially biomass

Then, energy of briquette --> 8000 BTU / lb

so..

mass of briquette = 93.18 g

change to lb,

93.18 g * (1 lb /454g) = 0.20524 lb of briquette

then

Energy from briquette = Heat Cap * Mass = 0.20524 lb * 8000 BTU/lbm = 1641.92 BTU

but only 80% transfers to water

so

Energy actually used for water = 80/100 * 1641.92 = 1313.536 BTU

then

T initial = 72F --> change to C

T(C) = (72-32)/1.8 C = 22.2222 C

Tfinal = 100 °C

Then... we will need to "boil" this so

Latent Heat water = 2264.76 kJ/kg

mass of water = M ( we don't know how much)

so

Qsensible heat = m*C*(Tf-Ti) = m*4.184 kJ/kg*C *(100-22.22)°C

Qlatent heat = m*LH = m*2264.76 kJ/kg

Qtotal = Qsensible + Qlatent = m*C*(Tf-Ti) + m*LH = m*4.184 kJ/kg*C *(100-22.22)°C + m*2264.76 kJ/kg

Qtotal = 1313.536 BTU --> change to kJ --> 1313.536 *

1 BTU = 1.055 kJ

so

Qtotal (KJ) = 1313.536 *1.055 = 1385.78048 kJ will heat...

1385.78048 kJ = m*4.184 kJ/kg*C *(100-22.22)°C + m*2264.76 kJ/kg

solve for m

m ((4.184*77.78) + 2264.76 ) = 1385.78048

m = 1385.78048 kJ /2590.19152 kJ/kg = 0.5350 kg

change to punds

0.454 kg = 1 lb

0.5350 kg = x lb

mass in lb = 0.5350/0.454 = 1.1794731 lbs of water

so...

Vol of water = 1.1794731 lb / 8.3 lb/gall = 0.14210 gallons