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 -10^{o}C to steam at 150^{o}C. Calculate the total energy required for the journey.

**The strings**:
S_{7}P_{7}A_{32} (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 1^{o}C.
**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 0^{o} (its melting point)

So energy required q = m(ΔT)C_{p} -------(1)

Where m is mass of ice; ΔT is change in temperature in degree Centigrade; C_{p} = specific heat of ice.

m = 10 g; ΔT = 10^{o}C; C_{p} = 2.06 Joules/g.C^{o}.

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 ΔH_{fusion} -------(2)

Where m is mass; ΔH_{fusion} = Enthalpy of fusion

m = 10 g; ΔH_{fusion 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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The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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