Anonymous
Anonymous asked in Science & MathematicsChemistry · 3 weeks ago

Thermodynamic Help!!?

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 2.20  kg of water to go from -19.0 °C to 73.0 °C.

fus-6.01 kj

cps-37.1 kj

cpl-75.3 j

1 Answer

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  • 3 weeks ago

    The "properties provided" are useless without their complete units.  They should be:

    fusion: 6.01 kJ/mol

    heat capacity, solid: 37.1 kJ/(kmol·°C)  [Who uses kilomoles for standard units?]

    heat capacity, liquid: 75.3 J/(mol·°C)

    (2.20 x 10^3 g H2O) / (18.01532 g H2O/mol) = 122.12 mol H2O

    Supposing the "water" at -19.0°C is actually ice:

    (37.1 kJ/(kmol·°C)) x (0.12212 kmol) x (0 - (-19.0))°C =

    86.082 kJ to warm the ice to its melting point

    (6.01 kJ/mol) x (122.12 mol) = 733.941 kJ to melt the ice

    (75.3 J/(mol·°C)) x (122.12 mol) x (73.0 - 0)°C = 671281 J =

    671.281 kJ to heat the melted ice to 73.0 °C

    86.082 kJ + 733.941 kJ + 671.281 kJ = 1491 kJ total

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