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worty [1.4K]
3 years ago
7

Find the initial temperature of Aluminum if 35 g of it produces 781.2 Joules at the final temperature of 44.3 ºC.

Chemistry
1 answer:
Annette [7]3 years ago
6 0

Answer:

T initial =  19.55°C

Explanation:

Given data:

Mass of Al = 35 g

Energy produces = 781.2 J

Final temperature = 44.3°C

Initial temperature = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of Al is 0.902 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

Q = m.c. ΔT

781.2 J = 35 g×0.902 J/g.°C× (T final - T initial)

781.2 J = 35 g×0.902 J/g.°C× (44.3°C -T initial )

781.2 J = 31.57 j/°C× (44.3°C -T initial )

781.2 J/ 31.57 j/°C = (44.3°C -T initial )

24.75°C = (44.3°C -T initial )

24.75° - 44.3°C = -T initial

T initial =  19.55°C

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A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
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A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

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From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

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            E_{a} = Activation Energy

             R = Universal constant

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taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
3 years ago
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