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Ray Of Light [21]
4 years ago
8

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J 2.80×107 J of

energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L 1.00 L of crude oil, if the water has its temperature raised from 23.5 °C 23.5 °C to 100 °C 100 °C , it boils, and the resulting steam is raised to 315 °C.
Chemistry
1 answer:
Alborosie4 years ago
4 0

Answer:

The question is incorrect and incomplete. Here's the correct question:

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J  of energy when burned. To illustrate this difficulty,a) calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L  of crude oil, if the water has its temperature raised from 23.5 °C to 100 °C , it boils, and the resulting steam is raised to 315 °C. b)Discuss additional complications caused by the fact that crude oil has less density than water.

Explanation:

Q= mc ΔT

Q= heat energy

m is mass

ΔT is change in temperature and c is specific heat capacity

calculating heat for latent heat of vaporisation

Q= ml where l is latent heat of vaporisation

a) Total heat energy used= heat required to raise temperature from 23.5 °C to 100 °C, heat required to boil water and heat required to further raise temperature from 100 °C to 315°C

Q = mc ΔT₁ + mL + mc ΔT₂

Q = m(c ΔT₁ + L + c ΔT₂)

m= Q÷(c ΔT₁ + L + c ΔT₂)

Q= 2.8 X 10⁷ J

c=4186J/kg°C

L=2256 x 10³J/kg

ΔT₁=76.5°C(100°C-23.5°C)

ΔT₂= 215°C(315°C-100°C)

(c ΔT₁ + L + c ΔT₂)= 4186J/kg°C *76.5°C + 2256 x 10³J/kg + 4186J/kg°C*215°C =3476219J/Kg

m= 2.8 x 10⁷J ÷3476219J/Kg

m =80.54 Kg

volume = mass÷ density

=80.54kg ÷ 10³kg/m³( density of water)

=0.0854m³

0.001m³ = 1 lL0.08054m³= 0.08054m³ /0.001m³= 80.54L

VOLUME is 80.54litres

b) since the density of crude is less than the density of water,and 80L of additional water is added, it'll make the crude to float on water thus inhibiting the extinguishing process

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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
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Answer:

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Explanation:

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k = Ae^{-\frac{E_{a}}{RT}}

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And for the reaction with the catalyst:

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Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

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\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
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