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

Here's how to calculate the total heat required for this process:<br /><br />**1. Heating the ice from -20°C to 0°C:**<br /><br />* Mass of ice (m) = 866 g<br />* Specific heat of ice (c_ice) = 2.03 J/g°C<br />* Temperature change (ΔT) = 0°C - (-20°C) = 20°C<br /><br />* Heat required (q1) = m * c_ice * ΔT <br />* q1 = 866 g * 2.03 J/g°C * 20°C = 35131.6 J = 35.13 kJ<br /><br />**2. Melting the ice at 0°C:**<br /><br />* Molar mass of water (M) = 18.015 g/mol<br />* Moles of ice (n) = m / M = 866 g / 18.015 g/mol = 48.07 mol<br />* Heat of fusion (ΔH_fus) = 6.02 kJ/mol<br /><br />* Heat required (q2) = n * ΔH_fus<br />* q2 = 48.07 mol * 6.02 kJ/mol = 289.3 kJ<br /><br />**3. Heating the water from 0°C to 100°C:**<br /><br />* Specific heat of water (c_water) = 4.184 J/g°C (This value is commonly used and more accurate than the one provided)<br />* Temperature change (ΔT) = 100°C - 0°C = 100°C<br /><br />* Heat required (q3) = m * c_water * ΔT<br />* q3 = 866 g * 4.184 J/g°C * 100°C = 362254.4 J = 362.25 kJ<br /><br />**4. Vaporizing the water at 100°C:**<br /><br />* Heat of vaporization (ΔH_vap) = 40.79 kJ/mol<br /><br />* Heat required (q4) = n * ΔH_vap<br />* q4 = 48.07 mol * 40.79 kJ/mol = 1960.11 kJ<br /><br />**5. Heating the steam from 100°C to 139°C:**<br /><br />* Specific heat of steam (c_steam) = 1.99 J/g°C<br />* Temperature change (ΔT) = 139°C - 100°C = 39°C<br /><br />* Heat required (q5) = m * c_steam * ΔT<br />* q5 = 866 g * 1.99 J/g°C * 39°C = 67557.54 J = 67.56 kJ<br /><br />**6. Total heat required:**<br /><br />* Total heat (q_total) = q1 + q2 + q3 + q4 + q5<br />* q_total = 35.13 kJ + 289.3 kJ + 362.25 kJ + 1960.11 kJ + 67.56 kJ = 2714.35 kJ<br /><br />Therefore, the total heat needed to convert 866 g of ice at -20°C to steam at 139°C is approximately $\boxed{2714}$ kJ.<br />