Question

How much heat energy, in kilojoules, is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer to three significant figures and include the appropriate units.

Answer 1

Answer:

25.2 kJ of heat energy is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C

Explanation:

Data: mass of ice = m = 53.0 g

         Temperature of ice = T1 = -18.0 ∘C

         Temperature of water = T2 = 25.0 ∘C

          Change in Temperature = ΔT : T2-T1                                          

           Specific heat of ice = c(i) = 2.09 J/g . ∘C

          Specific heat of water = c(w) = 4.18 J/g . ∘C

          Enthalpy of fusion of water (when convert from ice to water) = ΔH(f) : 334 J/g

          Enthalpy of vapourization of water (when convert from water to gas) = ΔH(v) :2250 J/g

          Total heat required = q = ?

Solution:        Heat required to melt the ice

                      T1 = -18 ∘C

                      T2 = 0 ∘C

                      Q1 = m x c(i) x ΔT

                      Q1 = 53.0 x 2.09 x (0-(-18))

                      Q1 = 53.0 x 2.09 x 18

                      Q1 = 1993.86 J is required by ice to reach to its

                      melting point.

                      Heat required to convert the ice into water

                      Q2 = ΔH(f) x m

                      Q2 = 334 x 53.0 = 17702 J is required by ice to  

                      convert into water.

                      Heat required by water to reach at 25 ∘C

                       T1 = 0 ∘C

                       T2 = 25 ∘C

                       Q3 = m x c(w) x ΔT

                       Q3 = 53.0 x 4.18 x (25-0)

                       Q3 = 5538.5 J is required by water to reach at  

                       25∘C .

                       Total heat required by 53.0 g of ice at −18.0 ∘C  

                        to water at

                        25.0 ∘C

                        q = Q1 + Q2 + Q3

                        q = 1993.86 J + 17702 J + 5538.5 J

                        q = 25234.36 J is the total heat required 53.0 g  

                        of ice at −18.0 ∘C to water at 25.0 ∘C.

                        Conversion from Joule (J) to Kilojoule (kJ)

                          1000 J =  1 kJ

                          q = 25234.36 J/1000

                          q = 25.23436 kJ

                           Conversion of heat in kJ to three significant  

                           figures

                           q = 25.2 kJ