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Masja [62]
3 years ago
8

When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th

e calorimeter increases by 10.0°C. The heat capacity of the calorimeter is 6.97 kJ/°C. Determine the energy flow, q (reaction).
Physics
1 answer:
balandron [24]3 years ago
7 0

Answer:

Q (reaction) = -69.7 kJ

Explanation:

Octane reacts with oxygen to give carbon dioxide and water.

C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O

This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C

The heat (q) of the reaction is calculated as follows:

Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature

q = -(6.97) x (10) = -69.7kJ

<em>Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required </em>

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              ∑ F = 0

 

Where the bold indicates vectors, F is the force and the sum is for all external forces.

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         2T_y - W = 0

Let's use trigonometry to decompose the tension, we can see from the graph and the adjoint that each string is half the length, let's call the angle θ

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The text indicates that the length of the block is 2 m, so the distance to the midpoint is

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Let's use the Pythagoras' Theorem            

            H² = CA² + CO²

           CO = \sqrt{H^2 - CA^2}

           CA = x

           CO = \sqrt{(\frac{L}{2} )^2  - 1 }

Where CO is the opposite leg,  CA is the adjacent leg and H is the hypotenuse indicating H = L / 2,

Let's write the trigonometry functions

           sin θ = \frac{CO}{H}

Let's substitute        

            sin θ = \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}

Let's subtitute in the equation  1

          2 T  ( \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}   ) = W

          \sqrt{\frac{L^2}{4}-1  }  = \frac{1}{4}  \ \frac{W}{T} \ L

Let's solve by squaring

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