At Saralstudy, we are providing you with the solution of Class 11 Physics Thermal properties of Matter according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 11 Physics Thermal properties of Matter so that you can prepare for the exam according to your own pace and your speed.
The given temperature, T = 27°C can be written in Kelvin as:
27 + 273 = 300 K
Outer diameter of the steel shaft at T, d1 = 8.70 cm
Diameter of the central hole in the wheel at T, d2 = 8.69 cm
Coefficient of linear expansion of steel, αsteel = 1.20 × 10–5 K–1
After the shaft is cooled using ‘dry ice’, its temperature becomes T1.
The wheel will slip on the shaft, if the change in diameter, Δd = 8.69 – 8.70
= – 0.01 cm
Temperature T1, can be calculated from the relation:
Δd = d1αsteel (T1 – T)
0.01 = 8.70 × 1.20 × 10–5 (T1 – 300)
(T1 – 300) = 95.78
∴T1= 204.21 K
= 204.21 – 273.16
= –68.95°C
Therefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.
(a) A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.
(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.
Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool. Thus, a brass tumbler feels colder than a wooden tray on a chilly day.
(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open.
Black body radiation equation is given by:
E = σ (T4 - T40)
Where, E = Energy radiation
T = Temperature of optical pyrometer
To = Temperature of open space
σ = Constant
Hence, an increase in the temperature of open space reduces the radiation energy.
When the same piece of iron is placed in a furnace, the radiation energy, E = σT4
(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth's surface.
(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).