A sample of water with a mass of 27.56g and an unknown temperature loses 2443 Joules. If the final temperature is found to be 62.50 C. What was the initial temperature?

Answer 1

q = mC∆T (memorize this equation)

q = heat = 2443 J

m = mass = 27.56 g

C = specific heat = 4.184 J/gº for water

∆T = change in temperature = ? = Tf - Ti = 62.50 - Ti

Solve for ∆T:

62.50 - Ti = (2443) / (27.56)(4.184)

62.50 - Ti = 21.2

Ti = 41.3º