Question

45.0g Pb at 115C is added to 125 g of water at 25.0C in a well insulated calorimeter. What is trhe final temperature in C for the water? The Specific heat of lead is 0.129J/gC and the specific heat of water is 4.184 J/gC.
Could someone please explain the proper way this equation can be arranged?
heat lost by Pb + heat gained by water = 0

heat lost by Pb = mass Pb x specific heat Pb x (Tfinal-Tinitial)

heat gained by H2O = mass H2O x specific heat H2O x (Tfinal-Tinitial)

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