If 335 g of water at 65.5 °C loses 9750 J of heat, what is the final temperature of the water? Liquid water has a specific heat of 4.18 J/(g×°C).

Answer 1

q = mC∆T

9750 J = (335 g)(4.18 J/gº)(∆T)

∆T = 9750 / (335)(4.18)

∆T = 6.9º

Final temperature of water = 65.5º - 6.9º = 58.6ºC