WBCHSE Class 11 Physics Transmission Of Heat Long Answer Questions

Unit 7 Properties Of Bulk Matter Chapter 9 Transmission Of Heat Long Answer Type Questions

Question 1. Two rods A and B are of equal length. Each rod has the ends at temperature T₁ and T₂ respectively, what is the condition that will ensure equal rates of flow of heat through the rods A and B?
Answer:

Given:

Two rods A and B are of equal length. Each rod has the ends at temperature T₁ and T₂ respectively

Let the cross-section and conductivity of rod A be α₁ and k₁ and those of rod B be α₂ and k₂ respectively.

Also, let the length of each rod be l.

Since, heat current in both rods are the same, \(\frac{Q}{t}=\frac{k_1 \alpha_1\left(T_1-T_2\right)}{l}\)

= \(\frac{k_2 \alpha_2\left(T_1-T_2\right)}{l}\)

or, \(k_1\alpha_1=k_2 \alpha_2\)

This is a required condition.

Read and Learn More Class 11 Physics Long Answer Questions

Question 2. At what temperature will a wooden and a metal block appear to be equally Hot or cold upon touching them?
Answer:

If the temperatures of the wooden block and of the metal block are the same as our body temperature, then they would appear to be equally hot or cold on touch.

Question 3. If a piece of paper Is wrapped around a wooden rod and held over a lighted candle the paper burns to ashes almost immediately. On the other hand, If the piece of paper Is wrapped on a metal rod and similarly held over the lighted candle then the paper takes some time to burn. Explain.
Answer:

Given:

If a piece of paper Is wrapped around a wooden rod and held over a lighted candle the paper burns to ashes almost immediately. On the other hand, If the piece of paper Is wrapped on a metal rod and similarly held over the lighted candle then the paper takes some time to burn.

Wood is a bad conductor of heat Obviously, the heat the wood receives from the candle is not conducted through it. So, the region that is heated retains the heat, as a result of which the paper reaches its ignition temperature and burns to ashes almost immediately.

If the wood is replaced by a metal rod, the heat is conducted through the latter, as the metal is a good conductor of heat. So, it takes some time for the paper to reach its ignition point and to start burning.

Question 4. Thermal conductivity of air is less than that of felt, still felt is more widely used as heat insulator. Why?
Answer:

Given:

Thermal conductivity of air is less than that of felt, still felt is more widely used as heat insulator.

Air has a thermal conductivity less than felt. Yet when a hot body is kept in air, a convection current develops. So the body starts losing heat. On the other hand, this is not the case if the body is covered with felt.

• This is because felt itself is a bad conductor of heat and the narrow perforations within the felt fibres are filled up with air. Air being a bad conductor of heat, no heat loss takes place due to conduction.
• Again, the body being covered with felt, it is not open to air. Hence, no heat loss takes place due to convec¬tion as well. So, the hot body can retain the heat. This is why felt is more widely used as a heat insulator.

Question 5. Why snow is a better insulator than ice?
Answer:

Snow is a better insulator than ice:

Ice being solid and having no pores do not enclose air. Snow being porous encloses air in it. Air is a bad conductor of heat hence snow is a better insulator.

Question 6. We feel wanner when the sky is cloudy—explain.
Answer:

Given:

We feel wanner when the sky is cloudy

Heat waves cannot penetrate cloud. Earth’s surface receives heat during daytime and radiates that out during night hours, thereby cooling the surface, Cloud acta m a reflector of this radiated float and sends It back to earth, So the cooling effect Is less and the nights are warm.

Question 7. A metal piece cooled In liquid air (temperature -180°C) produces a burning sensation when It is touched, why?
Answer:

Given:

A metal piece cooled In liquid air (temperature -180°C) produces a burning sensation when It is touched

The temperature of liquid air is -180°C l.e., much less than our body temperature. Hence, the temperature difference between the metal piece and our body is very large.

Metal Is a good conductor of heat, So, when the metal piece Is touched, heat conducts very quickly from the body to the metal piece. This very rapid loss of heat will produce a burning sensation, as our body cannot withstand It.

Question 8. Three specimens made of the same material are kept at 200°C In a room. They arc respectively a sphere, a cube, and a circular lamina of the same mass. Compare the rates of their cooling.
Answer:

Given:

Three specimens made of the same material are kept at 200°C In a room. They arc respectively a sphere, a cube, and a circular lamina of the same mass.

The mass, the temperature, and the material of the three specimens being the same, the rate of radiation from the surfaces is directly proportional to the surface areas. Hence, the circular lamina, having the largest surface area among the three, cools the fastest. The sphere, being of the smallest surface area, will have the slowest rate of cooling.

Question 9. How does a heater coll attain a steady temperature when current is on, after an Initial steady rise In temperature?
Answer:

Given:

Current through the coil of the heater produces heat and the coil shows a steady rise in temperature initially. With the increase in temperature rate of radiation from the coil also increases.

At a certain temperature, the rate of heating due to current equals the rate of radiation, i.e., the heater cannot retain any heat for a further rise in temperature. So, the coil maintains its constant temperature.

Question 10. Explain why the climate of a harbor town is more temperate than that of a town in a desert at the same latitude.
Answer:

Sea water has a high specific beat So the temperature of a harbour town does not increase very much due to heat absorption in the daytime, also if does not decrease very much due to the radiation of heat in bight time.

On the other hand desert sand has a low specific heat so due to heat absorption and radiation the temperature variation is much Higher, That is why (he climate of a harbor town is more temperate.

Question 11. A solid copper sphere of radius ft and a hollow sphere of the same material of external and internal radii R and r are raised to the same temperature and are allowed to cool under similar conditions. Which one of the two will have a faster rate of cooling?
Answer:

Given:

A solid copper sphere of radius ft and a hollow sphere of the same material of external and internal radii R and r are raised to the same temperature and are allowed to cool under similar conditions.

The external radius of the hollow sphere and the radius pf the solid sphere being the same, the outer surface areas of the two spheres are equal. If they are heated to the same temperature and are exposed in the same surroundings, the rate of loss of heat will be the same as their outer surface areas and Initial temperatures are equal.

But, for the same rate of heat loss, the temperature of the hollow sphere will drop at a faster rate as Its mass is less. So the hollow sphere will have a faster rate of cooling.

Question 12. Bulbs of two thermometers are coated with lampblack and sliver. Compare their readings when they are

1. Kept immersed In water inside a dark room,
2. In open daylight and
3. In a clear night.

Answer:

1. insulation is produced because of air being trapped between the bulb and the coating of lampblack. This will delay the thermometer to attain the temperature of water, Silver being a good conductor, the thermometer with the silver-coated bulb will attain thermal equilibrium with water instantaneously.

2. Lamp black is a good absorber and a bad reflector of heat. On the other hand silver is a good reflector and a bad absorber of heat. Hence the bulb coated with lamp black will absorb larger amount of heat in the sun¬light. On the other hand, the bulb coated with silver will reflect most of the heat. So, the first thermometer will show a higher reading.

3. In a clear cloudless night, the bulb coated with lamp black will radiate faster as it is a good absorber and hence a good radiator of heat. Hence, its temperature reading will be lower.

Question 13. Two spheres of the same material and of radii 1 m and 4 m are kept at 4000 K and 2000 K respectively, Showing that the heat radiated from them per second Is the same.
Answer:

We know that the relations among the rate of radiation (E), surface area (A), and absolute temperature (T) are \(E \propto A \text { and } E \propto T^4 \quad \text { or, } E \propto A T^4\)

∴ In case of the two spheres in question, \(\frac{E_1}{E_2}=\frac{A_1 T_1^4}{A_2 T_2^4}=\left(\frac{1}{4}\right)^2 \times\left(\frac{4000}{2000}\right)^4=\frac{1}{16} \times 16=1 \text { or, } E_1=E_2\)

Question 14. Water boils faster in a metal vessel with a rough and black bottom than in a metal vessel with a smooth bottom. Explain.
Answer:

Given:

Water boils faster in a metal vessel with a rough and black bottom than in a metal vessel with a smooth bottom.

The rough and dark-bottomed surface has a greater ability to absorb heat than the smooth-bottomed metal vessel. The smooth surface of the metal vessel reflects a reasonable part of the incident heat. Obviously, the water in the metal vessel with a rough and black bottom will boil faster.

Question 15. The bottom surface of a cooking utensil is made j rough and black, whereas a calorimeter surface is kept smooth and shiny. Explain.
Answer:

Given:

The bottom surface of a cooking utensil is made j rough and black, whereas a calorimeter surface is kept smooth and shiny.

The basic calorimetric principle (heat lost = heat gained) is easily applicable when no heat is exchanged between the calorimeter and its surroundings. The calorimeter surface is made bright and shiny to reduce heat exchange by radiation. But a rough and black-bottomed surface of a cooking utensil absorbs heat faster from the source.

Question 16. When we cover the bulb of a thermometer with a piece of quilt why is the reading of the instrument comparatively less? What changes in the reading will be recorded if the quilt is soaked?

1. In either or
2. In water?

Answer:

A quilt is a bad conductor of heat because of the wool and the air trapped within it. So the temperature of the surroundings cannot be attained by the bulb covered with a piece of quilt. Therefore, the reading of the thermometer is comparatively less.

1. If the quilt is soaked in ether, which is a highly volatile substance, the ether will evaporate quickly and take its latent heat of vaporization from the bulb of the thermometer. So, the reading in the thermometer will fall quickly.

2. If water is used instead of ether, it is less volatile, the reading in the thermometer will fall slowly.

Question 17. Two friends, waiting for their third companion in a restaurant, ordered two cups of tea. While one of them poured milk into his cup of tea, the other waited till their friend arrived and then mfwH the cold milk with the tea. Whose cup of tea would be comparatively warmer?
Answer:

Given:

Two friends, waiting for their third companion in a restaurant, ordered two cups of tea. While one of them poured milk into his cup of tea, the other waited till their friend arrived and then mfwH the cold milk with the tea.

The friend who mixed milk with the tea earlier would enjoy a warmer cup of tea. We know from Newton’s law of cooling, the more the difference in the temperature of the substance with its surroundings, more is the rate of heat radiation from it.

Since the first friend’s cup of tea became comparatively cooler on mixing the colder milk, the rate of heat radiation diminished in this case. Again, the second friend poured the cold milk afterward so, the rate of heat radiation from his tea was faster and hence it became colder

Question 18. What will be the nature of v_m T graph for a per fleetly black body?
Answer:

We know that for a perfectly black body

⇒ \(\lambda_m T=\text { constant }\)

or, \(\frac{T}{\nu_m}=\text { constant }\)

∴ \(\nu_m \propto T\)

So, v_m -T graph will be a straight line passing through the origin.

Question 19. Five rods of the same dimensions are arranged as shown. They have thermal conductivities k₁, k₂, k₃, k₄ and k₅. When points A and B are maintained at different temperatures, under what condition no heat flow through the central rod?

Answer:

The arrangement of the rods is similar to a bal¬anced Wheatstone Bridge.

No heat flows through the central rod if we have,

∴ \(\frac{k_1}{k_2}=\frac{k_3}{k_4} \text { or, } k_1 k_4=k_2 k_3\)

Question 20. The plots of intensity (I) versus wavelength (λ) for three black bodies at temperatures T₁, T₂, and T₃ respectively are as shown. What will be the relation of these temperatures?
Answer:

Given:

The plots of intensity (I) versus wavelength (λ) for three black bodies at temperatures T₁, T₂, and T₃ respectively are as shown.

According to Wein’s displacement law for blackbody radiation,

λ_m = constant

Therefore, we can conclude that the higher the value of λ_m, the smaller is the value of temperature.

According to the figure, λ₂ > λ₃ > λ₁

Therefore, T₁ > T₃ > T₂.

Question 21. A planet is situated at a mean distance d from the sun and its average surface temperature is T. Suppose that the planet receives energy only from the sun and loses energy from its surface by the process of radiation. Ignore all atmospheric actions. If \(T \propto d^{-n}\) then show that n = \(\frac{1}{2} \text {. }\)
Answer:

Given:

A planet is situated at a mean distance d from the sun and its average surface temperature is T. Suppose that the planet receives energy only from the sun and loses energy from its surface by the process of radiation.

Suppose, the power of radiation by the sun = P, radius of the planet = R

So, energy received by the planet = \(\frac{P}{4 \pi d^2} \times \pi R^2\)

Energy radiated by the planet = 4πR² • σT⁴

Therefore, for thermal equilibrium, \(\frac{P}{4 \pi d^2} \times \pi R^2=4 \pi R^2 \sigma T^4\)

or, \(T^4=\frac{P}{16 \pi \sigma d^2}\)

∴ \(T^4 \propto \frac{1}{d^2} \quad \text { or, } T \propto \frac{1}{d^{1 / 2}}\)

or, \(T \propto d^{-1 / 2}\)

Comparing with the given equation \(T \propto d^n \text {, we get } n=\frac{1}{2}\)