A container with $1 kg$ of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $700 Wm ^{-2}$ and it is absorbed by the water over an effective area of $0.05 m ^2$. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in ${ }^{\circ} C$ ) in the temperature of water and the surroundings after a long time will be. . . . . . . . (Ignore effect of the container, and take constant for Newton's law of cooling $=0.001 s ^{-1}$, Heat capacity of water $\left.=4200 J kg ^{-1} K ^{-1}\right)$
IIT 2020, Diffcult
Download our app for free and get started
$\frac{ dQ }{ dt }=\sigma e A \left( T ^4- T _0^4\right) $. . . . . $(i)$
$\left.\frac{d Q}{A d t}=e \sigma\left( T _0+\Delta T \right)^4- T _0^4\right)=\sigma T _0^4\left[\left(1+\frac{\Delta T }{ T _0}\right)^4-1\right]$
$=e \sigma T _0^4\left[\left(1+4 \frac{\Delta T }{ T _0}\right)-1\right]$
$\frac{ dQ }{ Adt }=\sigma e T _0^3 \cdot 4 \Delta T$ $. . . . .(ii)$
Now from equ. $(i)$
$ms \frac{ dT }{ dt }=\sigma e T \left( T ^4- T _0^4\right)$
$\frac{ dT }{ dt }=\frac{\sigma e A }{ ms }\left[\left( T _0+\Delta T \right)^4- T _0^4\right]$
$=\frac{\sigma e e }{ ms } T _0^4 \times\left[\left(1+\frac{\Delta T }{ T _0}\right)^4-1\right]$
$\frac{ dT }{ dt }=\frac{\sigma e A }{ ms } T _0^4-4 \Delta T$
$\frac{ dT }{ dt }=e \Delta T ;\left( K =\frac{4 \sigma e A T_0^s}{ ms }\right)$
$\Rightarrow 4 \sigma e A T_0^3=\frac{ K }{ A }( ms )$
from equ. $(1)$
$\frac{d Q}{A d t}=e \sigma I _0^3 \cdot 4 \Delta T$
$700=( K / A )( ms ) \Delta T$
$\therefore \Delta T =\frac{700 \times 5 \times 10^{-2}}{10^{-5} \times 4200}=\frac{50}{6}=\frac{25}{3}$
$\Delta T =8.33$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A sphere of density $\rho $ , specific heat capacity $c$ and radius $r$ is hung by a thermally insulating thread in an enclosure which is kept at a lower temperature than the sphere. The temperature of the sphere starts to drop at a rate which depends upon the temperature difference between the sphere and the enclosure and the nature of the surface of sphere and is proportional toView Solution
- 2View SolutionFollowing graph shows the correct variation in intensity of heat radiations by black body and frequency at a fixed temperature
- 3A black body is at a temperature of $2880\;K$. The energy of radiation emitted by this object with wavelength between $499\;nm$ and $500\;nm$ is ${U_1}$, between $999\;nm$ and $1000\;nm$ is ${U_2}$ and between $1499\;nm$ and $1500\;nm$ is ${U_3}$. The Wein's constant $b = 2.88 \times {10^6}\;nm\,K$. ThenView Solution
- 4Two rods of same length and cross section are joined along the length. Thermal conductivities of first and second rod are ${K_1}\,\,{\rm{and}}\,\,{K_2}$. The temperature of the free ends of the first and second rods are maintained at ${\theta _1}\,\,{\rm{and }}{\theta _2}$ respectively. The temperature of the common junction isView Solution
- 5If wavelengths of maximum intensity of radiations emitted by the sun and the moon are $0.5 \times {10^{ - 6}}m$ and ${10^{ - 4}}m$ respectively, the ratio of their temperatures isView Solution
- 6An object kept in a large room having air temperature of $25^{\circ} \mathrm{C}$ takes $12$ minutes to cool from $80^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$. The time taken to cool for the same object from $70^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ would be nearly.....$min$View Solution
- 7Two stars emit maximum radiation at wavelength $3600Å$ and $4800Å$ respectively. The ratio of their temperatures isView Solution
- 8Two rods having thermal conductivity in the ratio of $5 : 3$ having equal lengths and equal cross-sectional area are joined by face to face. If the temperature of the free end of the first rod is $100°C$ and free end of the second rod is $20°C$ . Then temperature of the junction is...... $^oC$View Solution
- 9View SolutionCan we boil water inside the earth satellite by convection
- 10Ice starts forming in lake with water at ${0^o}C$ and when the atmospheric temperature is $ - {10^o}C$. If the time taken for $1 \;cm$ of ice be $7$ hours, then the time taken for the thickness of ice to change from $1\; cm$ to $2\; cm$ isView Solution