Question

How much heat (in kJ) is needed to convert an 18.0-g cube of ice at 0.0 ^oC into liquid water at 20.0 ^oC?
For water (H₂O): heat of fusion = 6.02 kJ/mol, specific heat capacity (liquid) = 4.18 J/g ^oC.

           a)           1.61 kJ

b)           7.52 kJ

c)            89.6 kJ

d)           518 kJ

Answer 1

Answer

Q1= energy needed to phase change of H2O(from solid to liquid )

Q1= mass of H2O* latent heat of fusion of water

      = (18 g)*(334 J/g)= 6012 J

Q2= energy needed to warm the water from 0 to 20 degree C

Q2= mass of water *specific heat of water *temperature change

     =(18 g)*(4.18 J/gC)*(20)= 1504.8 J

Q= Q1+Q2= 6012+1504.8 = 7516.8 J= 7.52 kJ (option B)