e = 300g * 2.46 * (20 0)
q = m x Cg x (Tf Ti)
Cg = 2.44 J/(C g)
q = 300 g x 2.44 J/C g x (20 0) C = 14640 J
14.6 kJ
e = 14760 joules
= 14.76 kilo joules
First you need to look up the specific heat of ethanol. This value would be given on any standardized test.
e = 300g * 2.46 * (20 0)
now we just use the formula;
e=mc delta T
Where delta T = final temp. initial temp.
m = mass
c = specific heat
q = m x Cg x (Tf Ti)
Cg = 2.44 J/(C g)
q = 300 g x 2.44 J/C g x (20 0) C = 14640 J
14.6 kJ
e = 14760 joules
= 14.76 kilo joules
q = m x Cg x (Tf Ti)
Cg = 2.44 J/(C g)
q = 300 g x 2.44 J/C g x (20 0) C = 14640 J
14.6 kJ
This post is last updated on hrtanswers.com at Date : 1st of September – 2022