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baherus [9]
2 years ago
11

Explain the difference between heat and temperature. Does 1 L of water at 658°F have more, less, or the same quantity of energy

as 1 L of water at 65°C?
Chemistry
1 answer:
pogonyaev2 years ago
3 0

The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.

Heat is a type of energy that causes a person's body to feel hot or cold.

While the temperature of an object is a parameter that indicates how hot or cold the object is.

How is the temperature in degree Fahrenheit converted to degree celsius?

  • To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.

°C =( ^0F -32) * \frac{5}{9}

  • Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
  • Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-

°C = (658 F-32) *\frac{5}{9} = 347.78 \° C

  • Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.

To learn more about heat and temperature, visit:

brainly.com/question/20038450

#SPJ4

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Helppp ! Please :’)
Over [174]

Answer:

\boxed{\sf \lambda \:  = 1.282 \times    {10}^{ 3} nm}

Explanation:

Suppose an electron makes transition from n initial (ni) to n final (nf) then formula for wavelength is given by,

\frac{1}{ \lambda}  = R_H {Z}^{2}  \big(\frac{1}{{n_f^2}}  - \frac{1}{{n_i^2}} \big)

Where,

  • λ is wavelength of photon
  • Rʜ is rydberg constant, the value of Rʜ is

109690 Cm-¹ in <em>Puri Sharma Pathania</em> standard book of physical chemistry &

109737 Cm-¹ according to <em>Wikipedia</em>

  • Z is the atomic number of atom, for hydrogen Z =1,

& according to given data, ni = 5, nf = 3

Solution:

Let's solve for wavelength,

Substituting all the given data in above formula,

\frac{1}{ \lambda}  = 109690 \times  {1}^{2}  \big(\frac{1}{{3^2}}  - \frac{1}{{5^2}} \big)

\frac{1}{ \lambda}  = 109690 \times  {1}^{2}  \big(\frac{1}{{9}}  - \frac{1}{{25}} \big)

\frac{1}{ \lambda}  = 109690 \times  {1}^{2}  \times   \frac{25 - 9}{25 \times 9}

\frac{1}{ \lambda}  = 109690 \times  {1}^{2}  \times   \frac{16}{225}

\frac{1}{ \lambda}  =   7800.18 \: \: cm ^{ - 1}

\lambda = 1.282 \times  {10}^{ - 4} cm

Now we know that, 1 cm = 10000000 nm,

Converting the wavelength from Cm → Nm

\lambda \:  = 1.282 \times  {10}^{ - 4} \times  {10}^{7}

\lambda \:  = 1.282 \times  {10}^{ - 4 + 7}

\sf \lambda \:  = 1.282 \times    {10}^{ 3} nm

\sf \small Thanks \:  for \:  joining \:  brainly  \: community!

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Explanation:

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Since the number of protons and electrons must be equal in an atom for it to be neutral, if it looses electrons it becomes positively charged ion and if it gains electrons it becomes negatively charged ion.
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