Equal masses of paraffin wax and water are mixed in a container of negligible thermal capacity. If the initial temperature of water is 80oC and that of wax is 20oC, what is the final temperature of the mixture?

Explanation

We use the principle of conservation of energy, which states that the heat lost by the hotter substance (water) must equal the heat gained by the colder substance (paraffin wax), assuming no heat is lost to the surroundings.Qlost =Qgained The formula for heat transfer is Q=mcΔT, where:m is the massc is the specific heat capacityΔT is the change in temperatureFor this problem, we'll use the following values for specific heat capacity:Specific heat of water (cwater ) ≈ 4182 J/kg°CSpecific heat of solid paraffin wax (cwax ) ≈ 2140 J/kg°CSince the masses are equal (mwater =mwax ), we can set up the equation:mwater cwater ΔTwater =mwax cwax ΔTwax Because mwater =mwax , we can cancel out the mass variable:cwater ΔTwater =cwax ΔTwax Let Tf be the final temperature of the mixture. The change in temperature for each substance is:ΔTwater =(80−Tf )ΔTwax =(Tf −20)Now, substitute these into the equation:4182(80−Tf )=2140(Tf −20)Next, solve for Tf :334560−4182Tf =2140Tf −42800Group the temperature terms on one side and the constant terms on the other:334560+42800=2140Tf +4182Tf 377360=6322Tf Tf =6322377360 Tf ≈59.69°C