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The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, water, and steam are 2.09 j/g-k, 4.18 j/g-k, and 1.84 j/g-k, respectively. for h2o, hfus = 6.01 kj/mol, and hvap = 40.67 kj/mol

Respuesta :

Yna9141
First, calculate for the amount of heat used up for increasing the temperature of ice.

      H = mcpdT
       H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
    H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
    H = (18 g)(4.18 J/gK)(70 k)
    H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J

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