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2 years ago

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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22

.0ºC . The person has a normal skin temperature of 33.0ºC and a surface area of 1.50 m2. The emissivity of skin is 0.97 in the infrared, where the radiation takes place.

1 answer:

SIZIF [17.4K]2 years ago

3 0

Answer:

$5.45\times 10^{-4}$ W

Explanation:

$T_{r}$ = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

$T_{s}$ = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

$A$ = Surface area = 1.50 m²

$\epsilon$ = emissivity = 0.97

$\sigma$ = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

$R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})$

$R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})$

$R = 5.45\times 10^{-4}$ W

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(a) $3.77\cdot 10^5 MeV, 1.51\cdot 10^6 MeV, 3.39\cdot 10^3 GeV$

The energy levels of an electron in a box are given by

$E_n = \frac{n^2 h^2}{8mL^2}$

where

n is the energy level

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$m=9.11\cdot 10^{-31}kg$ is the mass of the electron

$L=1.0\cdot 10^{-15} m$ is the size of the box

Substituting n=1, we find the energy of the ground state:

$E_1 = \frac{1^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=6.03\cdot 10^{-8}J$

Converting into MeV,

$E_1 = \frac{6.03\cdot 10^{-8} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =3.77\cdot 10^5 MeV$

Substituting n=2, we find the energy of the first excited state:

$E_2 = \frac{2^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=2.41\cdot 10^{-7}J$

Converting into MeV,

$E_2 = \frac{2.41\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =1.51\cdot 10^6 MeV$

Substituting n=3, we find the energy of the second excited state:

$E_3 = \frac{3^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=5.43\cdot 10^{-7}J$

Converting into GeV,

$E_3 = \frac{5.43\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-9} GeV/eV =3.39\cdot 10^3 GeV$

(b) $1.10 \cdot 10^{-18} m$

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

$E=E_2 - E_1 = 2.41\cdot 10^{-7}J - 6.03\cdot 10^{-8} J=1.81\cdot 10^{-7} J$

And the energy of the electromagnetic radiation is

$E=\frac{hc}{\lambda}$

where c is the speed of light; so, re-arranging the formula, we find the wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.81\cdot 10^{-7}J}=1.10 \cdot 10^{-18} m$

(c) $6.59 \cdot 10^{-19} m$

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

$E=E_3 - E_2 = 5.43\cdot 10^{-7} J - 2.41\cdot 10^{-7}J =3.02\cdot 10^{-7} J$

Using the same formula as before, we find the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.02\cdot 10^{-7}J}=6.59 \cdot 10^{-19} m$

(d) $4.12 \cdot 10^{-19} m$

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

$E=E_3 - E_1 = 5.43\cdot 10^{-7} J - 6.03\cdot 10^{-8}J =4.83\cdot 10^{-7} J$

Using the same formula as before, we find:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.83\cdot 10^{-7}J}=4.12 \cdot 10^{-19} m$

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How much energy is required to take 500g of ice at -40 degrees to water at room temperature (22 degrees Celsius)? Show work pls

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Answer: Given that:

Mass (m) = 20 g = 0.02 Kg,

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Specific heat of water (Cp) = 4.187 KJ/Kg k

We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)

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Explanation:

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Romashka-Z-Leto [24]

The final volume of the gas is 0.75L

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Three children of masses and their position on the merry go round

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Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

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L(initial) = L(final)

Ii•ωi = If•ωf

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I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

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