If 40.0 g of water at 70.0°c is mixed with 40.0 g of ethanol at 10.0°c, what is the final temperature of the mixture?

When the heat lost by water = heat gained by ethanol

∴ (M * C * ΔT) w = (M*C*ΔT) eth

when Mw mass of water = 40 g

C specific heat of water = 4.18

ΔT = (70 - Tf)

and M (eth) mass (ethanol) = 40 g

C specific heat of ethanol = 2.44

ΔT = (Tf - 10)

by substitution:

40 * 4.18 * (70 - Tf) = 40 * 2.44 * (Tf-10)

∴Tf = 47.9 °C