Final temperature.......?

Question

Suppose a piece of lead with a mass of 14.9 g at a temperature of 92.5C is dropped into an insulated container of water. The mass of water is 165 g and its temperature before adding the lead is 20.0C. What is the final temperature of the system?

This is what I have so far...

q = (m) (T) (Cp) = (14.9g) (92.5C - Tf) (0.449 J / g * C)
q = (m) (T) (Cp) = (165g) (Tf - 20.0C) (4.184 J / g * C)

I have been working on this problem for 2 days now and I am having alot of trouble getting an answer. Help is highly appreciated!

Thanks so much!

Answer 1

You have the equations correct assuming the specifics hets you putin are right. SInce the "q" are changes in heat energy no the absolute value of heat energy, the heat lost by the lead is equal to the heat gained by the water. So set the two equations eqaul to each other:

mp*cpp*(Tp-Tf) = mw*cpw*(Tf -Tw)

where the "p" mans lead and the "w" means water (e.g. mp = mass of lead)

Thenn collect terms:

mp*cpp*TP +mw+cpw*Tw = Tf*(mw*cpw - mp*cpp)

so final temp is:

Tf = (mp*cpp*TP +mw+cpw*Tw )/(mw*cpw - mp*cpp)

Tf = (14.9*0.449*92.5 + 165*4.184*20)/(165*4.184-14.9*0.449)

Tf = 21.1 C