A sample of water with a mass of 35.75g and an initial temperature of 99.22 loses 12750 Joules. What is the final temperature? *

Answer 1

q = mC∆T

q = heat = 12750 J

m = mass = 35.75 g

C = specific heat = 4.184 J/gº (a constant for water that you can look up)

∆T = change in temperature = ?

Solve for ∆T:

∆T = q / (m)(C) = 12750 J / (35.75 g)(4.184 J/gº) = 85.24º

This is the CHANGE IN TEMPERATURE and since the water LOST energy (heat) the temperature will be lower than the initial temperature.

Final Temperature = 99.22º - 85.24º = 13.98º