[AhmedEzatAlzawalaty]

a problem says " i want to dry 100 kilos of a herb that contain 85 % miosture and the final product shall contain 5 % moisture , the initial temperature is 15 degrees and the final is 60.specific heat of water is 1 and specific heat of herb is 0.6 ,how can i calculate the heat required for drying?

[AhmedEzatAlzawalaty]

my solution is

In 100 Kilos of fresh herb mass of water is 100*85/100=85kg
mass of dry solid = 15 kg

for b kg of water in the product (dried solid) 100b/b+15=5
so water in the product is 0.789 kg
 
and so water to be removed by drying is 85-0.789=84.22 kg

heat required to heat 100 kg of herb =
for water 85*1*(60-15)=3825    (1)
for dry solid 15*0.6*45=405       (2)

amount of heat required to evaporate 84.22 kg of water =
84.22*540= 45478.8      (3)

NB the teacher says that the latent heat of vaporization of water is 540 but i found it on the web 2260 J/g (at 100oC)   
"i want to know the latent heat of vaporization of water at 60 degrees celsius" pleez

finally the total amount of heat required for drying will be summation of 1 and 2 and 3
will be 49708.8 k cal.

Am i being right?
if there is any explanation or comment or if something is missing or if i am wrong please any help will be greatly appreciated.

[AWK]

540 x 4.18 = 2257 ~2260
calories or Joules