Calculate the total heat change, the E, the PV work, and the H when 54 g of ice
ID: 690428 • Letter: C
Question
Calculate the total heat change, the E, the PV work, and the H when 54 g of ice is converted into54g of steam. [H2O(s) at -10°C converted toH2O(g) at 110°C]. Assume 1.0atm pressure. Use 2.0J/g °C for the specific heatsof ice and steam, and 4.2J/g °C for the specific heat of wateras a liquid. Use 6.0kJ/mole as the Hfus and41 kJ/mole as the Hvap for water. (warm ice+ melt ice + heat water + boil water + heat steam; sum the heatchanges for each step) Will ratelifesaver! Calculate the total heat change, the E, the PV work, and the H when 54 g of ice is converted into54g of steam. [H2O(s) at -10°C converted toH2O(g) at 110°C]. Assume 1.0atm pressure. Use 2.0J/g °C for the specific heatsof ice and steam, and 4.2J/g °C for the specific heat of wateras a liquid. Use 6.0kJ/mole as the Hfus and41 kJ/mole as the Hvap for water. (warm ice+ melt ice + heat water + boil water + heat steam; sum the heatchanges for each step) Will ratelifesaver!Explanation / Answer
Heat required for change of temperature = mass * specific heat *temperature diff . Heat required for change of state = mass * heat capacity . 54 g of ice from -10 C to 0 C: . Heat = 54 g * 2.0 J/g.C * (0-(-10)) C = 1080 J = 1.080 kJ . 54 g of ice at 0 C to water at 0 C moles of ice = mass/ molar mass = 54 g /18.02 g/mol = 3 moles . Heat = 3 moles * 6.0 kJ/mol = 18 kJ . 54 g of water from 0 C to 100 C Heat = 54 g * 4.2 J/g.C * (100-0) = 22680 J = 22.7 kJ . 54 g of water at 100 C to steam at 100 C Heat = 3 moles * 41 kJ/mol = 123 kJ . Steam at 100 C to steam at 110 C Heat = 54 g * 2.0 J/g.C * (110-100) C = 1080 J = 1.080 kJ . Total heat change = 1.080 kJ + 18 kJ + 22.7 kJ + 123 kJ + 1.080kJ =165.9 kJ
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