6th Edition Problem 8.21

[lizettelopez1F]

A piece of copper of mass 20.0 g at 100.0 C is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0 C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings.

What is the general equation that needs to be used to find the final temperature of the water?

[beckyolmedo1G]

heat lost by the copper= -heat gained by the water
so, you use the equation (mass of copper)(Tfinal-TInitial of copper)(C)=-(mass of water)(Tfinal-Tinitial of water)(C). Then solve for Tfinal

[KarlaArevalo2F]

hi becky, how did you derive this equation?

[Heesu_Kim_1F]

Hi, this might help.
The idea comes from Qsystem+Qsurr=0.
Qsystem = -Qsurr, in which copper = system and water = surrounding.
Thus, heat lost by copper = -heat gained by water.
Then use the calorimetry equation q = m x Cp x delta T to solve.

[Saman Andalib 1H]

The general concept that is used here is heat lost by the copper= -heat gained by the water. In the formula, the only thing that is a little bit tricky is that you have to use (Tfinal-Tinitial) for the variable delta T, then use algebra to simplify the equation to solve for Tfinal.

[Kobe_Wright]

What would cp be in this equation?

[Kobe_Wright]

What would cp be in this equation?

[Lynsea_Southwick_2K]

cp would be different for copper and water. Cp copper = o.38 J/Celsius*g and cp water is 4.18 J/Celsius*g

What would cp be in this equation?

[Chloe Likwong 2K]

What would cp be in this equation?

If the substance is anything except water, the Cp is usually given in the question.