What will be the final temperature when 15.0 kJ are added to 150 grams of water at 25°C?

Question

The partial answer is:4X°C

Answer 1

Heat is being added so the temperature will rise.

q = mC∆T

q = heat = 15.0 kJ

m = mass of water = 150 g = 0.150 kg

C = specific heat of water = 4.184 J/g/deg = 4.184 kJ/kg/deg

∆T = change in temperature = ?

∆T = q / (m)(C) = 15.0 kJ/(0.150 kg)(4.184 J/kg/deg)

∆T = 23.90 degrees = 24 degrees

Thus the final temperature will be 25ºC + 24ºC = 49ºC