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

Question

Following an example in the book I set the problem up as follows:
Heat lost by lead: (14.9g)(92.5 degreesC - Tf)(0.13J/g times C) and Heat gained by water: (165g)(Tf - 20.0 degrees C)(4.184J/g times C) but I am totally lost after this. I could not even get the answer on the example in the book. Please help!!!!!!!!
Thanks.

Answer 1

I don't see anything wrong with your work. I do it a little differently but both should work.
masslead x sp.h.lead x (Tf - Ti) + masswater x sp.h.water x (Tf-Ti) = 0
Tf is final T for which you solve.
Ti is initial T which is 92.5 for Pb and 20 for H2O.
Post your work if you still have trouble.
I get 20.2 C. That is such a small difference between the starting point and final point for water that it may be the number of decimals you carry your work before adding and subtracting. Check my work.