Option 4 : 16 P

__Concept:__

The radiation energy emitted by a body per unit time is given by:

Eb = ϵAσT4

Where ϵ is emissivity of the body.

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

__Calculation:__

For a given area E α T4

\(\frac{{{E_2}}}{{{E_1}}} = \frac{{T_2^4}}{{T_1^4}}\)

\(\frac{{{E_2}}}{{{E_1}}} = \frac{{{{\left( {2{T_1}} \right)}^4}}}{{T_1^4}} \Rightarrow {E_2} = 16E\),

E_{1} = P(given)

Option 3 : 4 : 1

__Concept:__

Rate of cooling is given by the expression ϵσA i.e. Q = ϵσ(πr2)

where ϵ is the emissivity, σ is the Stefan-Boltzmann constant = 5.67 × 10-8 W/m2K4, A = area, T = absolute temperature

The ratio of the diameters of two spherical balls is \(\frac{d_1}{d_2}=\frac{2}{1}\)

Let us assume the diameter of 1st ball is d1 = 2d and diameter of 2nd ball is d2 = d

So, the ratio of the rate of cooling of the big ball as compared to the smaller ball =

\(\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{{π d_1^2}}{{π d_2^2}} = {\left( {\frac{{2d}}{d}} \right)^2} = \frac{4}{1}\)

so, the ratio is 4 : 1

Option 3 : 81

__Concept:__

Stefan-Boltzmann Law

The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:

E ∝ T4

E = σT4

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

__Calculation:__

\(\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^4} = {\left( {\frac{{1227 + 273}}{{227\; + 273}}} \right)^4} = {\left( {\frac{{1500}}{{500}}} \right)^4} = 81\)

Option 4 : all of the above

__Explanation:__

The emissive power of a **Black Body** depends upon its temperature only, whereas in the case of a Grey Body it depends on emissivity as well as temperature.

Emissivity itself depends on

- Type of Material
- Physical nature of the surface
- Viewing angle
- Wavelength
- Geometry
- Temperature

Hence, we can say that the emissive power of a body depends upon all the above factors.

__Important Points__

The term "body" generally refers to "Grey Body", we cannot assume it "Black Body" unless or until it is specified.

\(E = \in \sigma {T^4}\)

\({E_b} = \sigma {T^4}\)

Option 4 : 81

__Concept:__

Stefan-Boltzmann Law

The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:

E ∝ T4

E = σT4

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

__Calculation:__

__Given:__

T_{1} = 400 K, T_{2} = 1200 K

\(\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4} = {\left( {\frac{{1200}}{{400}}} \right)^4} =3^4 = 81\)

A radiator in a domestic heating system operates at a surface temperature of 55°C. Assuming the radiator behaves as a black body, the rate at which it emits the radiant heat per unit area is (assume σ = 5.67 × 10^{-8} W/(m^{2}K^{4}))

Option 1 : 0.66 kW/m^{2}

**Concept:**

The radiant heat transfer is given as Q = σAT^{4}

where, σ = Stefan-Boltzmann constant. (5.67 × 10^{-8} W/m^{2}/K^{4}), A = area of body, T = absolute temperature

**Calculation:**

**Given:**

T = 55°C = 55 + 273 = 328 K, A = 1 m^{2}

⇒ 5.67 × 10^{-8} × (328)^{4}

= 656.26 W/m^{2}

= 0.656 kW/m^{2}

An electric flat-plate square heater of sides 10 cm provides 100 W power from each side. If the heater is assumed to be black, its temperature is approximately:

Option 2 : 648 K

**Concept:**

**Emissive power (E)** of a radiating body is given by:

**E = σAT ^{4} (in W) **

where σ = Stefan boltzmann's constant = 5.67 × 10^{-8} W/m-K^{4}, A = Surface area in m^{2}, T = Temperature of body(in K).

**Black body:**

A black body is a body which **absorbs radiations from all the directions** at all wavelength and emits the radiations above the particular temperature.

**Calculation:**

**Given:**

a = 10 cm = 0.1 m, P = 100 W, T = ?

Now, we know that

A black body absorbs radiations from all directions, So from the energy balance of the body

Power absorb = Emissive power (E) of body

**100** = σAT4

100 = 5.67 × 10^{-8} × 0.1 × 0.1 × T^{4}

**∴ T = 648 K**

Option 1 : 1702.9 kW

__Concept:__

Q = ϵσAT^{4}

Here A = surface area

∴ Surface area of cube = 6 × (side)^{2}

__Calculation:__

**Given:**

Side of cube = 10 m, T = 200° C = 473 K

**Now,**

Q = ϵσAT^{4}

∴ Q = 1 × 5.67 × 10^{-8} × 6 × (10^{2}) × (473)^{4}

**∴ Q = 1702 kW **

If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of

Option 4 : 16

**Concept:**

The Stefan-Boltzmann law for the emissive power gives the total energy emitted by a blackbody.

E_{b} = σT4

where Eb is the emissive power of a blackbody, T is absolute temperature, and σ (5.67 X 10-8 W/m2/K4) is the Stefan-Boltzmann constant.

**Calculation:**

**Given:**

T_{1} = T, T_{2} = 2T

So, \(\frac{E_{b2}}{E_{b1}}=(\frac{T_2}{T_1})^4\)

Therefore, \(\frac{E_{b2}}{E_{b1}}=(\frac{2T}{T})^4=16\)

Option 1 : 1 : 1.15

**Concept:**

The thermal energy emitted by a body is given as, Q = σAT^{4}

where, T is absolute temperature, and σ = (5.67 X 10-8 W/m2/K4) is the Stefan-Boltzmann constant.

From the given condition, we have,

\(\frac{{{Q_A}}}{{{Q_B}}} = \frac{{T_A^4}}{{T_B^4}}\)

**Calculation:**

**Given:**

T_{A} = 273 + 10 = 283 K, T_{B} = 273 + 20 = 293 K