Question
How much heat is needed to convert 866 g of ice at -20^circ C to steam at 139^circ C? (The specific heats of ice, steam, and liquid are 2.03J/gcdot ^circ C,1.99J/gcdot ^circ C, fusion is 6.02kJ/mol and heat of vaporization is 40.79kJ/mol. square x Your response differs from the correct answer by more than 10% Double check your calculations kJ Supporting Materials (1) Periodic Table Constants and Factors Supplemental Data
Solution
4.1
(253 Votes)
Brynn
Master · Tutor for 5 years
Answer
Here's how to calculate the total heat required for this process:**1. Heating the ice from -20°C to 0°C:*** Mass of ice (m) = 866 g* Specific heat of ice (c_ice) = 2.03 J/g°C* Temperature change (ΔT) = 0°C - (-20°C) = 20°C* Heat required (q1) = m * c_ice * ΔT * q1 = 866 g * 2.03 J/g°C * 20°C = 35131.6 J = 35.13 kJ**2. Melting the ice at 0°C:*** Molar mass of water (M) = 18.015 g/mol* Moles of ice (n) = m / M = 866 g / 18.015 g/mol = 48.07 mol* Heat of fusion (ΔH_fus) = 6.02 kJ/mol* Heat required (q2) = n * ΔH_fus* q2 = 48.07 mol * 6.02 kJ/mol = 289.3 kJ**3. Heating the water from 0°C to 100°C:*** Specific heat of water (c_water) = 4.184 J/g°C (This value is commonly used and more accurate than the one provided)* Temperature change (ΔT) = 100°C - 0°C = 100°C* Heat required (q3) = m * c_water * ΔT* q3 = 866 g * 4.184 J/g°C * 100°C = 362254.4 J = 362.25 kJ**4. Vaporizing the water at 100°C:*** Heat of vaporization (ΔH_vap) = 40.79 kJ/mol* Heat required (q4) = n * ΔH_vap* q4 = 48.07 mol * 40.79 kJ/mol = 1960.11 kJ**5. Heating the steam from 100°C to 139°C:*** Specific heat of steam (c_steam) = 1.99 J/g°C* Temperature change (ΔT) = 139°C - 100°C = 39°C* Heat required (q5) = m * c_steam * ΔT* q5 = 866 g * 1.99 J/g°C * 39°C = 67557.54 J = 67.56 kJ**6. Total heat required:*** Total heat (q_total) = q1 + q2 + q3 + q4 + q5* q_total = 35.13 kJ + 289.3 kJ + 362.25 kJ + 1960.11 kJ + 67.56 kJ = 2714.35 kJTherefore, the total heat needed to convert 866 g of ice at -20°C to steam at 139°C is approximately
kJ.