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