Specific Heat, Enthalpy Of Fusion, Enthalpy Of Vaporization
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Specific Heat, Enthalpy Of Fusion, Enthalpy Of Vaporization

There are five stages involved in the journey of 10 g of ice at -10oC to steam at 150oC. Calculate the total energy required for the journey.

The strings: S7P7A32 (force-push).

The math:
Pj Problem of Interest is of type force (force-push).

The concepts:
Specific Heat: the energy required to raise the temperature of 1 g of a substance by 1oC.
Enthalpy Of Fusion: the energy required to change 1 g of a substance from solid to liquid.
Enthalpy Of Vaporization: the energy required to change 1 g of a substance from liquid to gas.

Stage (a): temperature of ice is raised to 0o (its melting point)
So energy required q = m(ΔT)Cp -------(1)
Where m is mass of ice; ΔT is change in temperature in degree Centigrade; Cp = specific heat of ice.
m = 10 g; ΔT = 10oC; Cp = 2.06 Joules/g.Co.
So, q = 10(10)(2.06) = 206 J

Stage (b): the ice must be melted
So, energy required q = mass x enthalpy of fusion = m x ΔHfusion -------(2)
Where m is mass; ΔHfusion = Enthalpy of fusion
m = 10 g; ΔHfusion of ice = 334 J/g
So, q = 10 x 334 = 3340 J.

Stage (c): water must be heated from 0oC to its boiling point 100oC.
So using equation (1):
m = 10 g; ΔT = 100oC; Cp of water = 4.18 J/g.Co.
So, q = 10(100)(4.18) = 4180 J.

Stage (d): water must be vaporized
Enthalpy of vaporization of water = 2260 J/g
So by equation (2), q = 10 x 2260 = 22600 J

Stage (e): steam is then heated from 100oC to 150o. Cp of steam = 2.02 J/g.Co.
So, by equation (1), q = 10(50)(2.02) = 1010 J.

So, total energy required = 206J + 3340 J + 4180 J + 22600 J + 1010 J = 31, 300 J.

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