What is the enthalpy change during the process in which 100 g of liquid water at 50.0°C is cooled to ice at -30.0°C?cH2O(l) = 4.184 J/gꞏ°C cH2O(s) = 2.09 J/gꞏ°C Hfus = 6.01 kJ/molA. -60.6 kJB. -27.2 kJC. -13.6 kJD. -6.19 kJ **Understanding Enthalpy Change in Cooling Water**In this problem, we calculate the enthalpy change when 100 grams of liquid water at 50.0°C is cooled to ice at -30.0°C.### Given Data- **Specific heat of liquid water (\(c_{H2O(l)}\))**: 4.184 J/g°C- **Specific heat of ice (\(c_{H2O(s)}\))**: 2.09 J/g°C- **Enthalpy of fusion (\(\Delta H_{fus}\))**: 6.01 kJ/mol### Problem- Determine the enthalpy change for the process. ### OptionsA. -60.6 kJ B. -27.2 kJ C. -13.6 kJ D. -6.19 kJ To solve, consider the following steps:1. **Cool liquid water from 50.0°C to 0°C.**2. **Convert liquid water at 0°C to ice at 0°C.**3. **Cool ice from 0°C to -30.0°C.**These calculations involve changes in temperature and phase, measured by the specific heats and enthalpy of fusion provided.

Cooling of water from 50°C to 0°C Given: m = 100g ∆t = 0°C - 50°C = -50°C cH2O(l) = 4.184 J/g°C => q1 = mc∆T => q1 = (100g) × (4.184 J/g°C) × (-50°C) => q1 = - 20920 J => q1 = - 20.920 kJ Freezing water at 0°C ∆Hfus = 6.01 kJ/mol => q2 = - ∆Hfus × moles of water => q2 = - 6.01 × 100/18 => q2 = - 33.39 kJ Cooling of water from 0°C to -30°C Given : m = 100g ∆T = -30°C - 0°C = -30°C cH2O(s) = 2.09 J/g°C => q3 = mc∆T => q3 = 100 g × 2.09 J/g°C × (-30°C ) => q3 = - 6270 J => q3 = - 6.270 kJ Total heat energy released : q = q1 + q2 + q3 => q = (-20.92 kJ) + (-33.39 kJ) + (-6.270 kJ) => q = -60.6 kJ Hence, enthalpy change during process is equal to -60.6 kJ