23: Heat And Temperature : Neet-xii-physics

Home Learn Fast select Course

Change Unit Change Subject Last Menu

NEET-XII-Physics

23: Heat and Temperature

with Solutions - page 3

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

Qstn #3
A body A is placed on a railway platform and an identical body B in a moving train. Which of the following energies of B are greater than those of A, as seen from the ground?
(a) Kinetic
(b) Total
(c) Mechanical
(d) Internal
digAnsr: a,b,c
Ans : (a) Kinetic
(b) Total
(c) Mechanical
As body A is at rest on the ground, it possesses only potential energy, whereas body B, being placed inside a moving train, possesses kinetic energy due to its motion along with the train. Therefore, body B will have greater kinetic, mechanical (energy possessed by the body by virtue of its position and motion = kinetic energy+potential energy) energy and, hence, total (sum of all the energies) energy. No information is given about the temperature of the body so we can not say wheather body B^' s internal energy will be or will not be greater than that of body A.
Page No 12:

Qstn #4
In which of the following pairs of temperature scales, the size of a degree is identical?
(a) Mercury scale and ideal gas scale
(b) Celsius scale and mercury scale
(c) Celsius scale and ideal gas scale
(d) Ideal gas scale and absolute scale
digAnsr: c,d
Ans : (c) Celsius scale and ideal gas scale
(d) Ideal gas scale and absolute scale
Celsius scale and ideal gas scale measure temperature in kelvin (K) and the ideal gas scale is sometimes also called the absolute scale. A mercury scale gives reading in degrees and its size of degree, which depends on length of mercury column, doesn't match any of the above-mentioned scales.
Page No 12:

Qstn #5
A solid object is placed in water contained in an adiabatic container for some time. The temperature of water falls during this period and there is no appreciable change in the shape of the object. The temperature of the solid object
(a) must have increased
(b) must have decreased
(c) may have increased
(d) may have remained constant
digAnsr: a
Ans : (a) must have increased.
The whole system (water + solid object) is enclosed in an adiabatic container from which no heat can escape. After some time, the temperature of water falls, which implies that the heat from the water has been transferred to the object, leading to increase in its temperature.
Page No 12:

Qstn #6
As the temperature is increased, the time period of a pendulum
(a) increases proportionately with temperature
(b) increases
(c) decreases
(d) remains constant
digAnsr: b
Ans : (b) increases
In general, the time period of a pendulum,t, is given by
`` t=\frac{1}{2\pi }\sqrt{\frac{l}{g}}``.
When the temperature (T) is increased, the length of the pendulum (l) is given by,
`` l={l}_{0}(1+\alpha T)``,
where l₀ = length at 0 ^oC
`` \alpha ``= linear coefficient of expansion.
Therefore, the time period of a pendulum will be
`` t=\frac{1}{2\pi }\sqrt{\frac{{l}_{0}(1+\alpha T)}{g}}``
Hence, time period of a pendulum will increase with increase in temperature.
Page No 12:

#
Section : iv

Qstn #1
The steam point and the ice point of a mercury thermometer are marked as 80° and 20°. What will be the temperature on a centigrade mercury scale when this thermometer reads 32°?
Ans : Given:
Ice point of a mercury thermometer, T₀ = 20° C
Steam point of a mercury thermometer, T₁₀₀ = 80° C
Temperature on thermometer that is to be calculated in centigrade scale, T₁ = 32° C
Temperature on a centigrade mercury scale, T, is given as:
`` T\mathit{=}\frac{{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{0}}}{{T}_{\mathit{100}}\mathit{-}{T}_{\mathit{0}}}\times 100``
`` \Rightarrow T=\frac{32-20}{80-20}\times 100``
`` \Rightarrow T=\frac{12}{60}\times 100``
`` \Rightarrow T=\frac{120}{6}``
`` \Rightarrow T=20°\,\mathrm{\,C\,}``
Therefore, the temperature on a centigrade mercury scale will be 20^oC.
Page No 12:

Qstn #2
A constant-volume thermometer registers a pressure of 1.500 × 10⁴ Pa at the triple point of water and a pressure of 2.050 × 10⁴Pa at the normal boiling point. What is the temperature at the normal boiling point?
Ans : Given:
Pressure registered by a constant-volume thermometer at the triple point, P_tr = 1.500 × 10⁴ Pa
Pressure registered by the thermometer at the normal boiling point, P = 2.050 × 10⁴ Pa
We know that for a constant-volume gas thermometer, temperature (T) at the normal boiling point is given as:
`` T=\frac{P}{{P}_{tr}}\times 273.16\,\mathrm{\,K\,}``
`` \Rightarrow T=\frac{2.050\times {10}^{4}}{1.500\times {10}^{4}}\times 273.16\,\mathrm{\,K\,}``
`` \Rightarrow T=373.31\,\mathrm{\,K\,}``
Therefore, the temperature at the normal point (T) is 373.31 K.
Page No 12:

Qstn #3
A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water, find the melting point of lead.
Ans : Given:
In a gas thermometer, the pressure measured at the melting point of lead, P = 2.20 × Pressure at triple point(P_tr)
So the melting point of lead,(T) is given as:
`` T=\frac{P}{{P}_{tr}}\times 273.16\,\mathrm{\,K\,}``
`` \Rightarrow T=\frac{2.20\times {P}_{tr}}{{P}_{tr}}\times 273.16\,\mathrm{\,K\,}``
`` \Rightarrow T=2.20\times 273.16K``
`` \Rightarrow T=600.952\,\mathrm{\,K\,}``
`` \Rightarrow T\simeq 601\,\mathrm{\,K\,}``
Therefore, the melting point of lead is 601 K.
Page No 12:

Qstn #4
The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100°C)?
Ans : Given:
Pressure measured by a constant volume gas thermometer at the triple point of water, P_tr = 40 kPa = 40 × 10³ Pa
Boiling point of water, T = 100°C = 373.16 K
Let the pressure measured at the boiling point of water be P.
For a constant volume gas thermometer, temperature-pressure relation is given below:
`` T=\frac{P}{{P}_{tr}}\times 273.16\,\mathrm{\,K\,}``
`` \Rightarrow P=\frac{T\times {P}_{tr}}{273.16}``
`` \Rightarrow P=\frac{373.16\times 40\times {10}^{3}}{273.16}``
`` \Rightarrow P=54643\,\mathrm{\,Pa\,}``
`` \Rightarrow P=54.6\times {10}^{3}\,\mathrm{\,Pa\,}``
`` \Rightarrow P\simeq 55\,\mathrm{\,kPa\,}``
Therefore, the pressure measured at the boiling point of water is 55 kPa.
Page No 12:

Qstn #5
The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point.
Ans : Given:
Temperature of ice point, T₁ = 273.15 K
Temperature of steam point, T₂ = 373.15 K
Pressure of the gas in a constant volume thermometer at the ice point, P₁ = 70 kPa,
Let P_trbe the pressure at the triple point and P₂be the pressure at the steam point.
The temperature-pressure relations for ice point and steam point are given below:
For ice point,
`` {T}_{1}=\frac{{P}_{1}}{{P}_{tr}}\times 273.16\,\mathrm{\,K\,}``
`` ``
`` ``
`` \Rightarrow 273.15=\frac{70}{{P}_{tr}}\times {10}^{3}\times 273.16``
`` \Rightarrow {P}_{tr}=\frac{70\times 273.16\times {10}^{3}}{273.15}\,\mathrm{\,Pa\,}``
`` ``
For steam point,
`` {T}_{2}=\frac{{P}_{2}\times 273.16}{{P}_{tr}}\,\mathrm{\,K\,}``
`` ``
On substituting the value of P_tr ,we get:
`` 373.15=\frac{{P}_{2}\times 273.15\times 273.16}{70\times 273.16\times {10}^{3}}``
`` \Rightarrow {P}_{2}=\frac{373.15\times 70\times {10}^{3}}{273.15}``
`` \Rightarrow {P}_{2}=95.626\times {10}^{3}\,\mathrm{\,Pa\,}``
`` \Rightarrow {P}_{2}\simeq 96\,\mathrm{\,kPa\,}``
Therefore, the pressure at steam point is 96 kPa.
Page No 12:

Qstn #6
The pressures of the gas in a constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath, respectively. Find the temperature of the wax bath.
Ans : Given:
In a constant volume gas thermometer,
Pressure of the gas at the ice point, P₀ = 80 cm of Hg
Pressure of the gas at the steam point, P₁₀₀ = 90 cm of Hg
Pressure of the gas in a heated wax bath, P = 100 cm of Hg
The temperature of the wax bath `` \left(T\right)`` is given by:
`` T=\frac{P-{P}_{0}}{{P}_{100}-{P}_{0}}\times 100°\,\mathrm{\,C\,}``
`` \Rightarrow T=\frac{100-80}{90-80}\times 100``
`` \Rightarrow T=\frac{20}{10}\times 100``
`` \Rightarrow T=200°\,\mathrm{\,C\,}``
Therefore, the temperature of the wax bath is 200^o C.
Page No 12:

Qstn #7
In a Callender’s compensated constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel.
Ans : Given:
Volume of the bulb in a Callender's compensated constant pressure air thermometer, (V) =
1800 cc
Volume of mercury that has to be poured out, V' = 200 cc
Temperature of ice bath, T_o = 273.15 K
So the temperature of the vessel(T') is given by:
`` T\text{'}=\frac{V}{V-V\text{'}}\times {T}_{0}``
`` \Rightarrow T\text{'}=\frac{1800}{1600}\times 273.15\,\mathrm{\,K\,}``
`` \Rightarrow T\text{'}=307.293``
`` \Rightarrow T\text{'}\simeq 307\,\mathrm{\,K\,}``
Therefore, the temperature of the vessel is 307 K.
Page No 12:

Qstn #8
A platinum resistance thermometer reads 0° when its resistance is 80 Ω and 100° when its resistance is 90 Ω.
Find the temperature at the platinum scale at which the resistance is 86 Ω.
Ans : Given:
Resistance at 0^oC, R₀ = 80 `` \,\mathrm{\,\Omega \,}``
Resistance at 100^oC, R₁₀₀ = 90 `` \,\mathrm{\,\Omega \,}``
Let t be the temperature at which the resistance (R_t) is 86 `` \,\mathrm{\,\Omega \,}``.
`` t=\frac{{R}_{t}-{R}_{0}}{{R}_{100}-{R}_{0}}\times 100``
`` \Rightarrow t=\frac{86-80}{90-80}\times 100``
`` \Rightarrow t=\frac{6}{10}\times 100``
`` \Rightarrow t=60°``
Therefore, the resistance is 86 `` \,\mathrm{\,\Omega \,}`` at 60^oC.
Page No 13:

Qstn #9
A resistance thermometer reads R = 20.0 Ω, 27.5 Ω, and 50.0 Ω at the ice point (0°C), the steam point (100°C) and the zinc point (420°C), respectively. Assuming that the resistance varies with temperature as R_θ = R₀ (1 + αθ + βθ²), find the values of R₀, α and β. Here θ represents the temperature on the Celsius scale.
Ans : Given:
Reading on resistance thermometer at ice point, R₀ = 20 Ω
Reading on resistance thermometer at steam point, R₁₀₀ = 27.5 Ω
Reading on resistance thermometer at zinc point, R₄₂₀ = 50 Ω
The variation of resistance with temperature in Celsius scale,θ, is given as:
`` {R}_{100}={R}_{0}\left(1+\alpha \theta +\beta {\theta }^{2}\right)``
`` \Rightarrow {R}_{100}={R}_{0}+{R}_{0}\alpha \theta +{R}_{0}\beta {\theta }^{2}``
`` \Rightarrow {R}_{100}={R}_{0}+{R}_{0}\alpha \theta +{R}_{0}\beta {\theta }^{2}``
`` \Rightarrow \frac{\left({R}_{100}-{R}_{0}\right)}{{R}_{0}}=\alpha \theta +\beta {\theta }^{2}``
`` \Rightarrow \frac{27.5-20}{20}=\alpha \theta +\beta {\theta }^{2}``
`` \Rightarrow \frac{7.5}{20}=\alpha \times 100+\beta \times 10000...\left(i\right)``
`` \,\mathrm{\,Also\,},{R}_{420}={R}_{0}\left(1+\alpha \theta +\beta {\theta }^{2}\right)``
`` \Rightarrow {R}_{420}={R}_{0}+{R}_{0}\alpha \theta +{R}_{0}\beta {\theta }^{2}``
`` \Rightarrow \frac{{R}_{420}-{R}_{0}}{{R}_{0}}=\alpha \theta +\beta {\theta }^{2}``
`` \Rightarrow \frac{50-20}{20}=420\alpha +176400\beta ``
`` \Rightarrow \frac{3}{2}=420\alpha +176400\beta ...\left(ii\right)``
Solving (i) and (ii), we get:
α = 3.8 ×10^-3°C^-1
β = -5.6 ×10^-7°C^-1
Therefore, resistance R₀is 20 Ω and the value of α is 3.8 ×10^-3°C^-1and that of β is -5.6 ×10^-7°C^-1.
Page No 13:

Qstn #10
A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concrete is 1.0 × 10^-5 °C^-1.
Ans : Given:
Length of the slab when the temperature is 0°C, L₀ = 10 m
Temperature on the summer day, t = 35 °C
Let L₁ be the length of the slab on a summer day when the temperature is 35°C.
The coefficient of linear expansion of concrete, α = 1 ×10^-5 °C^-1
`` {L}_{1}={L}_{0}\left(1+\alpha t\right)``
`` \Rightarrow {L}_{1}``= 10 (1 + 10^-5 × 35)
`` \Rightarrow {L}_{1}`` = 10 + 35 × 10^-4
`` \Rightarrow {L}_{1}`` = 10.0035 m
So, the length of the slab on summer day when the temperature is 35^oC is 10.0035 m.
Page No 13: