If 62.7 joules of heat energy are added to 1.0 grams of water at 20° C, what will be the resulting temperature of the water?

Question

The Partial answer is 3X° C

Answer 1

q = mC∆T

q = heat = 62.7 J

m = mass of water = 1.0 g

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

∆T = change in temperature = ?

62.7 J = (1.0 g)(4.184 J/g/deg)(∆T)

∆T = 62.7 J/(1.0 g)(4.184 J/g/deg)

∆T = 15ºC

Final temperature = 20º + 15º = 35ºC