Question

A 400 gram sample of alcohol at 10o C [Ca=2.64 J/g*oC] is mixed with 400 grams of warm water [Cw=4.186 J/g*Co] at 88o C. Assume no heat is lost to the surroundings, what is the final temperature of the mixture when thermal equilibrium is reached?
Alcohol = -[Warm Water]
mCaΔT = -[mCwΔT]
can u help pls?

Answer 1

The heat lost by the water will be equivalent to the energy gained by the alcohol. Thus:
maCaΔT = -mwCwΔT
400 x 2.64 x (T - 10) = 400 x 4.186 x (88 - T)
T = 57.8 °C