Question

25.0 g of water cooled 10 C to -20 C. Calculate the total Energy change.

With these useful numbers: ΔH fus = 334 J/g Δvap = 2260 J/g C_p for ice = 2.08 J/g ⁰C C_p for liquid water = 4.18 J/g ⁰C C_p for steam = 1.8 J/g ⁰C

Answer 1

Q = heat change for conversion of water at 10^oC to water at 0 ^oC + heat change for conversion of water at 0^oC to ice at 0^oC + heat change for conversion of ice at 0 ^oC to ice at -20 ^oC

Amount of heat released , Q = mcdt + mL + mc'dt

                                          = m(cdt + L + c'dt' )

Where

m = mass of water = 25.0 g

c = Specific heat of water = 4.18 J/g degree C

c' = Specific heat of ice= 2.08 J/g degree C

L = Heat of fusion of ice = 334 J/g

dt = 10 -0 =10 ^oC

dt' = 0-(-20) = 20 ^oC

Plug the values we get Q = m(cdt + L + c'dt' )

                                      = 10.464 J

                                      = 10.44 kJ