If 4.5 g of water at 1.0° C absorb 33 joules of heat, what is the temperature change of the water?

Question

The partial answer is: 1.X°C

Answer 1

q = mC∆T

q = heat = 33 joules

m = mass of water = 4.5 g

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

∆T = change in temperature = ?

33 J = (4.5 g)(4.184 J/g)(∆T)

∆T = (33 J)/(4.5 g)(4.184 J/g/deg)

∆T = 1.8ºC (to 2 significant figures)