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Sonbull [250]
3 years ago
15

The element hydrogen has the highest specific heat of all elements. At a temperature of 25°C, hydrogen’s specific heat capacity

is 14300J/(kg K). If the temperature of a .34kg sample of hydrogen is to be raised by 25 K, how much heat will have to be transferred to the hydrogen?
Physics
1 answer:
natali 33 [55]3 years ago
8 0

Answer:

121550 J

Explanation:

Parameters given:

Mass, m = 0.34kg

Specific heat capacity, c = 14300 J/kgK

Change in temperature, ΔT = 25K

Heat gained/lost by an object is given as:

Q = mcΔT

Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:

Q = 0.34 * 14300 * 25

Q = 121550J or 121.55 kJ

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Unpolarized light of intensity Io is incident on a stack of 7 polarizing filters, each with its axis rotated 17°cw with respect
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The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

To answer the question, we need to know what polarization of light is.

<h3>What is polarization of light?</h3>

This is when the electric field vector of light is oscillating in one plane.

  • Now for light of intensity I' which is initially unpolarized, its intensity after polarization is I = 1/2I'.
  • Also, for light initially polarized, its intensity after polarization is I"' = I"cos²Ф where Ф is the angle between the initial direction and the direction of polarization.
<h3>Intensity of light through each polarized filter</h3>

Given that we have 7 polarizing filters, each rotated 17° cw with respect to the previous filter.

So, since the light is initially unpolarized,

  • The intensity through the first polarizing filter is I₁ = 1/2I₀ where I₀ is the initial intensity.
  • The intensity through the second polarizing filter is I₂ = I₁cos²17°= 1/2I₀cos²17°
  • The intensity through the third polarizing filter is I₃ = I₂cos²17° = 1/2I₀cos⁴17°
  • The intensity through the fourth polarizing filter is I₄ = I₃cos²17° = 1/2I₀cos⁶17°
  • The intensity through the fifth polarizing filter is I₅ = I₄cos²17° = 1/2I₀cos⁸17°
  • The intensity through the sixth polarizing filter is I₆ = I₅cos²17° = 1/2I₀cos¹⁰17°
  • The intensity through the seventh polarizing filter is I₇ = I₆cos²17° = 1/2I₀cos¹²17°.
<h3>The ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity</h3>

Since I₇ is the last intensity I₇ = It = 1/2I₀cos¹²17°.

So, It/I₀ = 1/2cos¹²17°

= 1/2(0.9563)¹²

= 1/2 × 0.5850

= 0.2925

So, the ratio of the intensity between light intensity that emerges from the last filter and unpolarized light of intensity, I₀ is It/I₀ = 0.2925

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The radius of curvature of the proton's path while in the field is 66.67  × 10^{-2}.

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