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charle [14.2K]
3 years ago
11

Calculate the amount of heat (in kj) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 k to 385.0 k. th

e specific heat capacity of ethanol is 2.42 j/g°c.
Physics
1 answer:
earnstyle [38]3 years ago
7 0
We are given with the specific heat capacity of ethanol, the mass of the sample and the temperature change to determine the total amount of heat to raise the temperature. The formula to be followed is H = mCpΔT. Upon subsituting, H = 79 g * 2.42 J/gC *(385-298)C = 16.63 kJ 
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Answer:

When a disaster is declared, the Federal government, led by the Federal Emergency Management Agency (FEMA), responds at the request of, and in support of, States, Tribes, Territories, and Insular Areas and local jurisdictions impacted by a disaster.

Explanation:

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2 years ago
Rami uses a disposable camera that has a flash. When he wants to take a picture, he holds a button and hears a rising whining so
ivann1987 [24]
<span>It stores energy and delivers it in a short burst.

The whirring sound is produced by the charging of the capacitor. A capacitor is an electrical component which is capable of storing charge. When the capacitor stores charge, it is storing energy. After doing so, the capacitor releases the electrical energy that it had stored as light energy, which is seen as the flash of the camera. It must do so in a burst, because the intensity of the flash is very high and would require a high amount of energy to maintain.
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3 years ago
Exercise will not help maintain the health of your endocrine system.
Mashcka [7]
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6 0
2 years ago
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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