At 1 bar, how much energy is required to heat 83.0 g of H_2 O(s) at...

Question

Question

At 1 bar, how much energy is required to heat 83.0 g of H20(s) at-14.0 oC to H20 (g) at 145.0 C? Number kJ

Answer 1

Answer #1

First,heat absorbed in raising the temperature of the ice to 0C from -14C

q1 = mCdeltaT

q1= 83×2.06J/gC×14C = 2393.72J = 2.39kJ

q2 = heat absorbed in melting ice at 0C to liquid water

q2 = (83g/18g/mol) × 6.02kJ/mol = 27.76kJ ( heat of fusion of water = 6.02kJ/mol)

q3 = heat absorbed in raising the temperature from 0C to 100C

q3 = 83g×4.184J/gC×100C = 34.73kJ

q4= heat absorbed in boiling water from 100C to steam at 100C

q4 = (83g/18g)× 40.7kJ/mol = 187.67kJ (heat of vaporization of water = 40.7kJ/mol)

q5 = heat absorbed in raising the temperature from 100C to 145C

q5 = 83g×2.02J/gC×45C = 7.54kJ

Q = q1+q2+q3+q4+q5

Q = 2.39+27.76+34.73+187.67+7.54 = 260.1kJ

Answer 2

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