14 questions · timed · auto-graded
Is nonuniform.
Explanation:
In steady state, the temperature of the rod is nonuniform maximum at the end near the furnace and minimum at the end that is away from the furnace.
Explanation:
When a hot liquid is kept in a big room, the liquid will loose its temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid.
As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease.
Therefore, the plot of temperature with time is best represented by the curve in option (a).
Explanation:
The thermal conductivity of a rod depends only on the material of the rod. For example, metals are much better conductors than non-metals because metals have large number of free electron that can move freely anywhere in the body of the metal and carry thermal energy from one place to other.
Also, 2 copper rods having different lengths and areas of cross-section have same thermal conductivity that depends only on the number of free electrons in copper.
Refract radiation.
Explanation:
A black body is an ideal concept. A black body is the one that absorbs all the radiation incident on it.
So, a black body does not reflect and refract radiation.
Convection currents.
Explanation:
Convection current is the movement of air (or any fluid) due to the difference in the temperatures. During summer days, there is temperature difference of air above the land and river. Due to this, a convection current is set from the river to the land during daytime. On the other hand, during night, a convection current is set from the land to the river.
Therefore, a mild air always flows on the shore of a calm river due to the convection current.
A straight line.
Explanation:
When a hot liquid is kept in a big room, then the liquid will loose temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid.
As the temperature decreases, the rate of loss will also decrease. So, the slope of the curve will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid. So, in steady state, the difference between the temperatures of the two will become zero.
A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The logarithm converts the fourth power dependence into a linear dependence with some coefficient (property of log). So, the plot satisfying all the above properties will be a straight line.
Stefan's law.
Explanation:
From Stefan-Boltzman's law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
$\text{u}=\sigma\text{AT}^4$
Where $\sigma$ is Stefan's constant.
Suppose a body at temperature T is kept in a room at temperature T0.
According to Stefan's law, energy of the thermal radiation emitted by the body per unit time is.
$\text{u}=\text{e}\sigma\text{AT}^4$
Here, e is the emissivity of the body.
The energy absorbed per unit time by the body is (due to the radiation emitted by the walls of the room)
$\text{u}_0=\text{e}\sigma\text{AT}^4_0$
Thus, the net loss of thermal energy per unit time is.
$\triangle\text{u}=\text{u}-\text{u}_0$
$\triangle\text{u}=\text{e}\sigma\text{A}(\text{T}^4-\text{T}_0^4)\ \dots(1)$
Newton law of cooling is given by,
$\frac{\text{dT}}{\text{dt}}=-\text{bA}(\text{T}-\text{T}_0)$
This can be obtained from equation (i) by considering the temperature difference to be small and doing the binomial expansion.
$\text{T}_2-\text{T}_1$
$\text{T}_2^2-\text{T}_1^2$
$\text{T}_2^3-\text{T}_1^3$
$\text{T}_2^4-\text{T}_1^4$
$\text{T}_2^4-\text{T}_1^4$
Explanation:
From stefan-Boltzmann law, the energy of thermal radiation per unit time by a blackbody of surface are A is given by,
$\text{u}-\sigma\text{AT}^4$
Here, $\sigma$ is stefan-Boltzmann constant.
Since the temperature of the solid is less than the surrouncings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The rate of emission from the soild will be proportional to $\text{T}_1^4$ and rate of emission from the surroundings will be proportional to $\text{T}_2^4.$
So, the net rate of increase in temperature will be proportional to $\text{T}_2^4-\text{T}^4_1.$
By all the three modes.
Explanation:
In conduction, heat is transferred from one place to other by vibration of the molecules. In this process, the average position of a molecule does not change. Hence, there is no mass movement of matter.
In convection, heat is transferred from one place to other by actual motion of particles of the medium. When water is heated, hot water moves upwards and cool water moves downwards.
In radiation process, transfer of heat does not require any material medium. For a room containing air, heat can be transferred via radiation (no medium required) and convection (by the movement of air molecules) and by conduction (due to collision of hot air molecules with other molecules).
1 : 1.15
Explanation:
From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
$\text{u}=\sigma\text{AT}^4$
Here, $\sigma$ is Stefan-Boltzmann constant.
The thermal radiation emitted in a given time by A and B will be in the ratio.
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{\text{T}_\text{A}^4}{\text{T}_\text{n}^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{(273+10)^4}{(273+20)^4}$
$\frac{\text{u}_\text{A}}{\text{u}_\text{B}}=\frac{1}{1.15}$
Explanation:
In steel, conductivity is higher than charcoal. So, if both are picked up by bare hands, then heat transfer from the body (steel or charcoal) to our hand will be larger in case of steel. Hence, steel will be hotter than the charcoal.
On the other hand, emissivity of charcoal is higher as compared to steel. So, if the two are picked up from the lawn and kept in a cold chamber, charcoal will lose heat at a faster rate than steel.
Explanation:
Let the temperature of the surroundings be T0.
From the Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by,
$\text{u}=\sigma\text{AT}_4$
Here, $\sigma$ is Stephen's constant.
Also, the energy absorbed per unit time by the body is given by,
$\text{u}_0=\text{e}\sigma\text{AT}_0^4$
As the two spheres have equal radii and temperatures, their rate of absorption and emission will be equal in the beginning.
More than 5 minutes.
Explanation:
Let the temperature of the surrounding be T°C.
Average temperature of the liquid in first case = 62.5°C
From newton law of cooling,
$1^\circ\text{C}\ \text{min}^-1=-\text{b}\text{A}(62.5-\text{T})^\circ\text{c}$
$\Rightarrow-\text{bA}=\frac{1}{62.5\text{T}}\text{min}^-1\ \dots(1)$
From Newton's law of cooling and equation (1),
$5^\circ\text{C}=-\text{bA}(57.5-\text{T})^\circ\text{C}$
$\Rightarrow\frac{5^\circ\text{C}}{\text{t}}=\frac{1}{62.5-\text{T}}(57.5-\text{T})^\circ\text{C}$
$\Rightarrow\text{t}=\frac{5(62.5-\text{T})}{(57.5-\text{T})}$
$\text{t}>5\ \text{minutes}.$
All bodies.
Explanation:
From Stefan-Boltzmann law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
$\text{u}=\sigma\text{AT}^4$
Here, $\sigma$ is Stefan-Boltzmann constant.
This law holds true for all the bodies.