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erastovalidia [21]
3 years ago
8

Which best describes cosmic microwave background radiation? radiation thought to exist but not yet measured radiation predicted

to exist in millions of years radiation that existed before the big bang leftover radiation from the big bang
Physics
1 answer:
grigory [225]3 years ago
8 0

Answer:  leftover radiation from the big bang

Microwave background radiation is a form of electromagnetic radiation that is present in the whole universe and completely fills it.

Its frequency belongs to the microwaves range and is one of the main demonstrations of the of the Big Bang theory model.

It is important to note the Big Bang occurred 400,000 years before the events related to the microwave cosmic radiation, this means the Big Bang occurred first.

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A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water
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Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer (\dot Q), measured in watts, that is, joules per second, by the following formula:

\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t} (1)

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m - Mass of the sphere, measured in kilograms.

c - Specific heat of the material, measured in joules per kilogram-degree Celsius.

T_{o}, T_{f} - Initial and final temperatures of the sphere, measured in degrees Celsius.

\Delta t - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}} (2)

Where:

m_{I}, m_{X} - Masses of the iron and unknown spheres, measured in kilograms.

\Delta t_{I}, \Delta t_{X} - Times of the iron and unknown spheres, measured in seconds.

c_{I}, c_{X} - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}

If we know that \Delta t_{I} = 6.35\,s, \Delta t_{X} = 4.59\,s, m_{I} = 0.515\,kg, m_{X} = 1.263\,kg and c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}, then the specific heat of the unknown material is:

c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)

c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

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