As the temperature differences are small, we can use Newton's law of cooling.

$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)\frac{d\theta }{\theta -{\theta }_0}=-kdt$                                                                          ...(i)
Where k is a constant,$\theta$ is the temperature of the body at time t and ${\theta }_0$ =  ${16}^{\circ }C$ is the temperature of the surrounding. We have,

${\int }_{{40}^{0}c}^{{36}^{0}c}\frac{d\theta }{\theta -{\theta }_0}=-k\left(5min\right)In\frac{{36}^{0}c-{16}^{0}c}{{40}^{0}c-{16}^{0}c}=-k\left(5min\right)$

or, $k=-\frac{In\left(5/6\right)}{5min}$

or, If t be the time required for the temperature to fall from 36⁰C to 32⁰C then by (i),

${\int }_{{36}^{0}c}^{{32}^{0}c}\frac{d\theta }{\theta -{\theta }_0}=-ktIn\frac{{32}^{0}c-{16}^c}{{36}^{0}c-{16}^0}=-\frac{In\left(5/6\right)t}{5min}t=\frac{In\left(4/5\right)}{In\left(5/6\right)}\times 5min=6.1min$

From the graph it is clear that initially both the bodies are at same temperature but after that at any instant temperature of body x is less than the temperature of body y. It means body x emits more heat i.e., emissivity of body x is more than body y $\therefore e_x>e_y$ and according to Kirchhoff's law good emitter are also good absorber so $a_x>a_y$.

According to Wein's displacement law $\lambda m\times T$ = Constant 

Here $\lambda m<\lambda m2<\lambda m1$                                                   $\Rightarrow T_3>T_2>T_1$

The temperature of Sun is higher than that of welding arc which in turn is greater than tungsten filament. 
Therefore $T_3$ is the Sun's temperature, $T_2$ the welding arc's and $T_1$ the tungsten filament's.

The inner surface of B radiates at a rate of $\frac{\Delta Q}{\Delta t}=\sigma A_B{T}_A^4$,

where                  $\sigma \to$ stefan's constant

AB $\to$Area

TB $\to$ Temperature

Similarly, A radiates $\frac{\Delta Q_2}{\Delta t}=\sigma A_B{T}_A^4$

Now, all the radiation from inner surface of B goes inside the hollow of the shell, as B is a very poor conductor. So, this radiation falls on A, and A being a blackbody absorbs all the radiation falling on it! And it being a good conductor the heat travels to the centre almost instantly, thus making the temperature uniform throughout the ball. Similarly, all the heat radiated by A is completely absorbed by inner surface of B, and none of it is lost to the surrounding as B is a very poor conductor.

∴ Net heat lost by B = Net heat gained by A =$\sigma (A_B{T}_B^4-A_A{T}_A^4)$

                                                                 = $5.67\times {10}^{-8}\times (80\times 293.1s^4-20\times 373.1s^4)\times {10}^{-4}$

                                                                 = 1.15 w

Now this is equal to specific heat $\times \frac{dtemperature}{dTime}$ of both A and B

$\therefore \frac{d{T}_A}{dt}={\frac{1.15}{42}}^{\circ }c/s={0.03}^{\circ }c/s$

$\frac{d{T}_B}{dt}={\frac{1.15}{82}}^{\circ }c/s={0.01}^{\circ }c/s$

Stefan's law: $\frac{\Delta Q}{\Delta t}=e\sigma A{T}^4$

For an ideal radiator e = 1                                                            $\Rightarrow \frac{\Delta Q}{\Delta t}=6\times {10}^{-8}\times 1.6\times (273+37{)}^4$

=$6\times {10}^{-8}\times 1.6\times 314\times 104$

= 887 J per second.

'Aluminum foil like' thermal blankets are designed to reflect most of the radiation, or infrared radiation.

Also, they are shiny! This makes them easy to spot from helicopters etc, in case of rescue operations.

So, all in all, a must have for campers!

Let the temperature of the coil be T. The coil will emit radiation at a rate A$\sigma T^4$. Thus,

1000W =$\left(0.020m^2\right)\times (6.0\times {12}^{-8}W/m^2-K^4)\times T64$

or,                                       $T^4=\frac{1000}{0.020\times 6.00\times {10}^{-8}}{K}^4$

                                               =$8.33\times {10}^{11}{K}^4$

or,                                            $T=955K$.

For all modes of heat transfer $\Delta Q\propto Area$,A.So we should choose the shape with minimum area for minimum heat transfer.

(a)For cubical:

Volume, V = $a^3=1l$,$\therefore a=1dm\left(decameter\right)$

Area, A = $6a^2=6d{m}^2$

(b)For conical: V= $\frac{\pi r^{2}h}{3}$, but given r=h,$r=\frac{3\frac{1}{3}}{\pi }$ dm

A=$\pi r\left(\sqrt{2r}\right)+\pi r^2$

= 7.354 $d{m}^2$

(c)For Sphere:

$V=\frac{4}{3}\pi r^3$, $\therefore r=(\frac{3}{4}{)}^{\frac{1}{3}}dm$

A$=4\pi r^2=4.84d{m}^2$

(d) For Cylinder:

$V=\pi \times r^2\times r\therefore r=\frac{1}{\pi \frac{1}{3}}dm$

A=$2\pi r^2+2\pi r.r=5.86d{m}^2$

As Sphere has the least area among these, it will be the shape with minimum heat loss. In fact sphere is the shape with the minimum surface area among all possible shapes!

The energy radiated per second by a black body is given by Stefan's Law as  $\frac{E}{t}=\sigma T^2\times A$ where A is the surface area of the black body

$\frac{E}{t}\sigma T^4\times 4\pi r^2$

Since black body is a sphere, $(A=4\pi r^2)$.

Case(i)

$\frac{E}{t}=450,T=500K,r=0.12m,$

$\therefore 450=4\pi \sigma \left(500{)}^4\right(0.12{)}^2$                              .....(i)

Case(ii)

$\frac{E}{t}=?,T=1000K,r=0.06m,$

dividing (ii) and (I) we get,

$\frac{\frac{E}{t}}{450}=\frac{\left(1000{)}^4\right(0.06{)}^2}{\left(500{)}^4\right(0.12{)}^2}=\frac{2^4}{2^2}=4$

$\Rightarrow \frac{E}{t}=450\times 4=1800W$

(d) is the correct option.

As the temperature differences are small, we can use Newton's law of cooling.

$\frac{d\theta }{dt}=-kt(\theta -{\theta }_0)$

$\frac{d\theta }{\theta -{\theta }_0}=-kdt$

or,                                               ...(i)

Where k is a constant, $\theta$ is the temperature of the body at time t and ${\theta }_0={16}^{\circ }C$ is the temperature of the surrounding. We have,

${\int }_{{40}^{\circ }C}^{{36}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt\left(5min\right)$

$ln\frac{{36}^{\circ }C-{16}^{\circ }C}{{40}^{\circ }C-{16}^{\circ }C}=-kt\left(5min\right)$

or,

                                     $k=-\frac{ln\left(\frac{5}{6}\right)}{5min}$

or, .

If t be the time required for the temperature to fall from 36${}^{\circ }$C to 32${}^{\circ }$C then by (i),

${\int }_{{36}^{\circ }C}^{{32}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt$

or,        $ln\frac{{32}^{\circ }C-{16}^{\circ }C}{{36}^{\circ }C-{16}^{\circ }C}=-\frac{ln\left(\frac{5}{6}\right)}{5min}$

or,          t=$\frac{ln\left(\frac{4}{5}\right)}{\left(ln\right(\frac{5}{6})}\times 5min$

= 6.1 min.

Alternative method

The mean temperature of the body as it cools from 40${}^{\circ }$C to 36${}^{\circ }$C is  $\frac{{40}^{\circ }C+{36}^{\circ }C}{2}={38}^{\circ }C$ The rate of decrease of temperature is

  $\frac{{40}^{\circ }C+{36}^{\circ }C}{5min}={0.80}^{\circ }C/min$ .

Newton's law of cooling is

$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$

or, -0.8${}^{\circ }$C/min = - k(38${}^{\circ }$C - 16${}^{\circ }$C)=-k(22${}^{\circ }$C)

or, $k=\frac{0.8}{22}mi{n}^{-1}$.

Let the time taken for the temperature to become 32${}^{\circ }$C be t.

During this period,.

The mean temperature is

Now,