Question

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?

Answer 1

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