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25 December, 22:23

Two 20.0-g ice cubes at - 17.0 C are placed into 205 g of water at 25.0C. Assuming no energy is transferred to or from the surroundings. Calculate the final temperature of the water after all the ice melts. Heat Capacity of H2O (s) 37.7 J / (mol. K) Heat Capacity of H2O (l) 75.3 J / (mol. K) Enthalpy of fusion of H2O 6.01 kJ/mol

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  1. 25 December, 22:28
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    The final temperature of the water is 6.53 °C

    Explanation:

    Step 1: Data given

    Mass of ice cubes = 2*20.0 grams = 40.0 grams

    Temperature of the ice = - 17.0 °C

    Mass of water = 205.0 grams

    Temperature of water = 25.0 °C

    Heat Capacity of H2O (s) = 37.7 J / (mol. K)

    Heat Capacity of H2O (l) = 75.3 J / (mol. K)

    Enthalpy of fusion of H2O 6.01 kJ/mol

    Step 2: Calculate moles of H2O

    Moles H2O = mass H2O / molar mass

    40.0 g H2O (s) / 18.02 g/mol = 2.22 mol H2O

    Step 3: Calculate energy needed to warm the ice to melting temperature

    (37.7 J / (mol*K)) * (2.22 mol) * (0 - (-17.0)) K = 1422.8 J

    Step 4: Calculate energy needed to melt the ice

    (6.01 KJ/mol) * (2.22 mol) = 13.34 kJ = 13340 J to melt the ice

    Step 5: Calculate total energy to warm and melt all the ice

    1422.8 + 13340 J = 14762.8 Joule

    Step 6: Calculate moles of water

    Moles H2O = 205 grams / 18.02 g/mol

    Moles H2O = 11.38 mol

    Step 7: Calculate the change in temperature

    ΔT = (14762.8 J / 75.3 J/mol*k) / 11.38 mol

    (15101.8 J) / (75.3 J / (mol*K) / (205 g / (18.01532 g H2O/mol)) = 17.2

    Step 8: Calculate new temperature of water

    25.0 - 17.2 = 7.8 °C

    We have 2 bodies of water: 40 g at 0°C and 205 g at 7.8°C

    ⇒ ((40 g * 0°C) + (205 g * 7.8°C)) / (40 g + 205 g) = 6.53 °C

    The final temperature of the water is 6.53 °C
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