A beaker contains 345.0 g of ethanol at 18.2°C. If the ethanol absorbs 1125 J of heat without losing heat to the surroundings, what will be the final temperature of the ethanol?

Question

(The specific heat of ethanol is 2.46 J/g·°C.)

Answer 1

q = mC∆T

q = heat = 1125 J

m = mass = 345.0 g

C = specific heat = 2.46 J/gº

∆T = change in temperature = ?

Solving for ∆T...

∆T = q / mC = 1125 J / (345.0 g)(2.46 J/gº)

∆T = 1.33º

Final temperature = 18.2º + 1.33º = 19.5º