Question

Calculate the standard enthalpy change for the phase transition reaction of solid water (ice) at -10°C into liquid water at 25°C if the enthalpy of fusion of ice is 6.01 kJ/mol and the specific heat capacity of ice and liquid water are 2.09 J/g°C and 4.18 J/g°C, respectively.

Answer 1

To calculate the standard enthalpy change for the phase transition reaction of solid water  ice  at -10C into liquid water at 25C, we need to consider three steps:1. Heating the ice from -10C to 0C2. Melting the ice at 0C3. Heating the liquid water from 0C to 25CFirst, let's calculate the enthalpy change for each step:1. Heating the ice from -10C to 0C:q1 = mass  specific heat capacity of ice  temperature changeq1 = m  2.09 J/gC   0 -  -10   Cq1 = 20.9 m J  where m is the mass of water in grams 2. Melting the ice at 0C:q2 = moles  enthalpy of fusionq2 =  m/18.015  mol  6.01 kJ/molq2 = 0.3335 m kJ3. Heating the liquid water from 0C to 25C:q3 = mass  specific heat capacity of liquid water  temperature changeq3 = m  4.18 J/gC   25 - 0  Cq3 = 104.5 m JNow, let's add the enthalpy changes for all three steps to find the total standard enthalpy change:Total enthalpy change = q1 + q2 + q3Total enthalpy change = 20.9 m J + 0.3335 m kJ + 104.5 m JTotal enthalpy change =  20.9 + 104.5  m J + 0.3335 m kJTotal enthalpy change = 125.4 m J + 0.3335 m kJTo convert the enthalpy change in kJ, divide by 1000:Total enthalpy change = 0.1254 m kJ + 0.3335 m kJTotal enthalpy change = 0.4589 m kJSo, the standard enthalpy change for the phase transition reaction of solid water  ice  at -10C into liquid water at 25C is 0.4589 m kJ, where m is the mass of water in grams.