Calculate the equilibrium temperature of the coffee.

Question

14.0g of cream at 11.5°C are added to an insulated cup containing 150.0g of coffee at 84.6°C. 

Calculate the equilibrium temperature of the coffee.

You may assume no heat is lost to the cup or surroundings, and that any physical properties of cream and coffee you need are the same as those of water.

Be sure your answer has the correct number of significant digits.

This question is different than previous questions and I'm not sure how to approach this.

Answer 1

Heat gained by the cream MUST equal heat lost by the coffee.

Heat gained by the cream = q = mC∆T = (14.0 g)(4.184 J/gº)((Tf - 11.5º) where Tf = final temperature

Heat lost by water = q = mC∆T = (150.0 g)(4.184 J/gº)(84.6º - Tf)

Set the two equations equal to each other and we have...

(14.0 g)(4.184 J/gº)((Tf - 11.5º) = (150.0 g)(4.184 J/gº)(84.6º - Tf)

Since 4.184 is constant on both sides (the C for water), we can eliminate that and end up with ..

(14.0 g)(Tf - 11.5º) = (150.0 g(84.6º - Tf)

14Tf - 161 = 12690 - 150Tf

164Tf = 12851

Tf = equilibrium temperature = 78.4ºC