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Inga [223]
2 years ago
9

Automobile engines contain cooling systems that transfer thermal energy from the engine to the environment. Without a cooling sy

stem, how would
the temperature of the engine be affected as its operating speed increases?

A The temperature would increase
B The temperature would decrease
C The temperature would stay the same
Chemistry
1 answer:
vovangra [49]2 years ago
6 0

When the engine's <em>operating speed increases</em> the temperature of the engine will also increase.

Given:

Automobile engines without a cooling system.

To find:

The affect on the temperature of the engine as its operating speed increases.

Solution:

The speed of the operating engine increases.

  • As we know that kinetic energy of the object and motion of the object is directly proportional<em> </em>to each other.

<em>Velocity </em><em>of object </em><em>∝ Kinetic energy </em><em>of object...[i]</em>

  • When the<em> kinetic energy</em> of the object increases the kinetic energy of the particles of the object also increases.
  • And <em>kinetic energy</em> of the particles is directly linked to the <em>temperature </em>of the object.

<em>Kinetic energy</em><em>  ∝ </em><em>Temperature </em><em>of object</em>...[ii]

From [i] and [ii] :

<em>Velocity </em><em>of object ∝  </em><em>Temperature </em><em>of an object</em>

  • When the engine's operating speed increases it will increase the kinetic energy of its particles and with that increase, the increase in the temperature of the engine will be <em>observed</em>.

So, from this, we can conclude when the engine's operating speed increases the temperature of the engine will also increase.

Learn more about kinetic energy and temperature here:

brainly.com/question/2731193?referrer=searchResults

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(a) Calculate the energy released by the alpha decay of 222 86Rn.
Alinara [238K]

Answer:

a). The energy released by the alpha decay of 222 Rn is to 218 Po :

= 5.596 MeV

b).The energy of the alpha particle is 5.5 MeV

c) The recoil energy of Po = 0.096 MeV

Explanation:

The equation for the alpha decay of Rn to Polonium is:

_{86}^{222}\textrm{Rn}\rightarrow _{2}^{4}\textrm{He}+_{84}^{218}\textrm{Po}

The energy of the decay process can be calculated using:

\Delta E=\Delta mc^{2}

\Delta m = Change in the mass

Mass of Po = 218.008965 u

Mass of He = 4.002603 u

Mass of Rn = 222.017576 u

\Delta m = mass of Po + mass of He - mass of Rn

=  4.002603 + 218.008965-222.017576  

= - 0.006008

\Delta E=m(931.5MeV)

\Delta E=-0.006008\times 931.5MeV

= -5.596 MeV

<u><em>The negative sign means energy is released during the process.</em></u>

b) The energy of the alpha particle is :

\frac{Po\ mass }{Rn\ mass}\times E

\frac{218.0089}{222.0175}\times \Delta E

\frac{218.0089}{222.0175}\times 5.596

= 5.494 MeV

= 5.5 MeV

c).

Recoil energy: When the parent nucleus is at rest before the decay then , there must be the some recoil of daughter nucleus to conserve the momentum.This is termed as recoil energy.

<u><em>The energy of the recoil polonium atom :</em></u>

The formula for recoil energy is :

<em>The total energy - the kinetic energy</em>

<em>= 5.596 - 5.5 </em>

<em>= 0.096 MeV</em>

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Which of the following pure compounds will exhibit hydrogen bonding?
zimovet [89]

Explanation:

For a compound to show hydrogen bonding it is necessary that the hydrogen atom of the compound should be attached to more electronegative atom like fluorine, oxygen or nitrogen.

For example, CH_{3}CH_{2}OH, CH_{3}NH_{2} and NH_{3} all these compounds contain an electronegative atom attached to hydrogen atom.

Therefore, these pure compounds will exhibit hydrogen bonding.

Thus, we can conclude that out of the given options CH_{3}CH_{2}OH, CH_{3}NH_{2} and NH_{3} are the pure compounds which will exhibit hydrogen bonding.

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