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At 1 atm, how much energy is required to heat 43.0 g of H2O(s) at –20.0 °C to H2O(g)...

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user7904696 | eNotes Newbie

Posted November 21, 2013 at 7:48 PM via web

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At 1 atm, how much energy is required to heat 43.0 g of H2O(s) at –20.0 °C to H2O(g) at 135.0 °C?

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jerichorayel | College Teacher | (Level 1) Senior Educator

Posted November 22, 2013 at 2:51 AM (Answer #1)

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First, we have to consider that there are three phases in this problem. Water exists in three phases: solid, liquid and gas. We started by heating the solid water at -20 degrees Celsius up to 0 degrees Celsius. At 0 degrees Celsius, the solid water or ice starts to melt. So the transformation is from solid water at 0 degrees Celsius to liquid water at 0 degrees Celsius. The next is the energy needed to as the water at 0 degrees is converted to water at 100 degrees. At 100 degrees the third transformation occurs. The liquid water at 100 degrees is converted into gaseous water at 100 degrees Celsius. Finally, the gaseous water is heated up to 135 degrees Celsius.  

In summary, the energy needed to heat the solid ice at -20 degrees Celsius to 169 degrees Celsius is:

q water (from -20 to 0) + heat of fusion  of water + q water (from 0 to 100) +heat of vaporization + q water (from 100 to 135)

Heat of fusion = 334.16 J g^-1
Heat of vaporization = 2259 J g^-1
specific heat capacity for solid water (ice) = 2.06 J g^-1 ) C^-1
specific heat capacity for liquid water = 4.184 J g^-1 C^-1
specific heat capacity for gaseous water (steam) = 2.02 J g^-1 C^-1

  • q water (from -20 to 0) = m Cp Delta T = 43 (2.06)(0 - (-20)) = 1594.44 J
  • heat of fusion of water = 334.16 J g^-1 * (43) = 14368.88 J
  • q water (from 0 to 100) = m Cp Delta T = 43 (4.184) (100-0) = 17991.2 J
  • heat of vaporization = 2259 J g^-1 * (43) = 97137 J
  • q water (from 100 to 135) = 43 (2.02) (135-100) = 3040.1 J

And then add everything...

= 1594.44 + 14368.88 + 17991.2 + 97137 + 3040.1 = 134131.62 J = 134.13 kJ of heat

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