Solution-Class-11-Science-Physics-Chapter-7-Thermal Properties of Matter-Maharashtra Board

(B) 34 ºC to 42 ºC

(D) Water contracts on freezing

(B) 45°

(B) α : β : γ = 1 : 2 : 3

(A) I and II are both correct

(C) 0.46 °C

• A thermometer is calibrated to measure a specific temperature on an empirical scale. A standard temperature interval is selected between two fixed points, the ice and steam points.
• The interval is divided into equal intervals, each called a degree of temperature,
• This procedure sets up an empirical scale for temperature.

There are three different scales of temperature namely the Celsius scale, Fahrenheit scale and absolute or thermodynamic scale. The units of temperature in these scales are the degree Celsius (°C), degree Fahrenheit (°F) and kelvin (K).

The relation between these three scales of temperature is

\(\frac{T_C}{100}=\frac{T_F-32}{180} =\frac{T_K-273.15}{100}\) 

OR 

\(\frac{T_C}{5}=\frac{T_F-32}{9} =\frac{T_K-273.15}{5}\)

The absolute zero of temperature, or the zero of the absolute or thermodynamic scale of temperature, is that temperature at which the energy of atoms and molecules of every material is minimum.

Since t (in °C) = T (in K) − 273.15, T = 0 K is

t = 0 K − 273.15 = − 273.15 °C

Also, in degree Fahrenheit,

t - 32 = \(\frac{9}{5}\)(T − 273.15) =  \(\frac{9}{5}\)(0 K − 273.15) = − 491.67 °F

∴ t = − 491.67 + 32 = − 459.67 °F

We have, β = 20: and γ = 3

∴ 3β = 6∝ ……..(1)

and 2γ = 6∝  ……..(2)

Combining Eqs. (1) and (2), we get,

6∝ = 3β = 2γ

Also, β = 2∝ and γ = 3∝. Hence, the relation between or, ∝, β and γ  is

∝ = \(\frac{β}{2}\)  = \(\frac{γ}{3}\)

Thus, the three coefficients are in the ratio

∝ : β : γ = 1 : 2 : 3

Thermal expansion joints are small gaps between rails on a railway track, created due to thermal expansion during summer and friction during train travel.

• Thermal expansion joints are used in railway tracks, bridge spans, and factories to allow rails and pipelines to expand in summer due to friction and heat.
• These joints are also used in bullock carts, where an iron rim is heated to expand to a larger circumference.
• A bimetallic strip with different expansion coefficients is used as a circuit breaker.
• A tightly fitted metal lid on a glass jar can be loosened by inserting it in hot water, which expands the lid.
• Thermometers use mercury and alcohol to measure temperature changes by measuring the volume expansion of these substances.

These joints help ensure the safety and efficiency of various applications, such as transportation, manufacturing, and medical applications.

If Q is the amount of heat supplied to a substance of mass 111 and AT is the rise in

temperature of the substance, the specific heat of the substance is

s = \(\frac{Q}{mΔT}\)

When a gas is heated isothermally,

Q ≠ 0 but ΔT = 0.

∴ s = ∞

On the other hand, if a gas is compressed or allowed to expand adiabatically, i.e., no heat enters or leaves the gas, its temperature changes

but Q = 0    ∴ s = 0

Thus, it can be seen that the specific heat capacity of a gas can have any value from 0 to ∞, depending upon the conditions under which it is heated. Therefore, to fix the value of the specific heat capacity of a gas, in practice, either the volume or the pressure is kept constant. These two being the most important or principal conditions under which a gas is usually heated, we define two principal specific heat capacities for a gas, namely,

(i) specific heat capacity at constant volume (s_v) and

(ii) specific heat capacity at constant pressure (S_p).

At a given pressure, the transition between liquid and solid states occur at the same temperature, i.e., the fusion or freezing point is the same as the melting point.

Sublimation : Sublimation is the direct phase transition from solid to gaseous state without passing through the intermediate liquid state.

Triple point : Triple point is the point at which the fusion curve, vaporization curve and sublimation curve meet. It is the only combination of temperature and pressure at which all the three phases of matter are in equilibrium and can therefore exist simultaneously.

Steady state : A thick metal rod of uniform cross section has one of its ends, ‘A’ maintained at a constant temperature by steam, see below Fig.

The opposite end ‘B’ is maintained at a constant lower temperature. The rod is covered with a thermal insulator (such as felt, slag wool or glass fibres — such materials are then called lagging) to prevent any heat loss from its longer surface. Thermometers are placed in holes drilled into the rod, with a small amount of mercury in each hole to ensure good thermal Contact between the rod and the bulb of a thermometer.

• Initially, as the end A is heated by steam, all the thermometers show a gradual rise in temperature.
• The section nearest to the steam end A always remains hottest.
• At this stage, as heat is being conducted, every section of the rod retains a part of this heat (which raises its temperature) and transmits the rest to the next section.
• After some time, the thermometer readings become stable, indicating that there is no further rise in the temperature of any section of the rod. Then, the rod is said to be in a thermal steady state.

Definition: The rate at which heat is conducted through two parallel surfaces of a block in the steady state divided by the area ofcross section and the temperature gradient is called the thermal conductivity or coefficient of thermal conduction of the material.

Let CG be a small section of length x of a lagged rod in a thermal steady state. The stable temperature at the end C is T1 and that at the end G is T₂ < T₁. The lagging ensures that there is no loss of heat to the surroundings through the cylindrical surface of the rod. Hence, all the heat passes through the rod in a direction normal to the cross sections C and G, (Fig.)

Since the rod is in a steady state, the quantity of heat entering per unit time through the end C of the section must be equal to that leaving the end G. It is found experimentally that the quantity of heat Q that flows from hot end to the cold end is directly proportional to

(i) the time t

(ii) the cross-sectional area A

(iii) the temperature difference between the two ends (T₁ — T₂) and is inversely proportional to

(iv) The perpendicular distance x between hot and cold ends That is,

Q ∝ A\((\frac{T_1-T_2}{x})\)t

Q = kA\((\frac{T_1-T_2}{x})\)t

where k is a constant of proportionality and is called the thermal conductivity. Thus, the thermal conductivity can be written as,

k = \(\frac{Qx}{A(T_1-T_2)t}\)

Applications of thermal conductivity:

(1) Thermal conductivity is a crucial factor in various applications, such as heating tubings in air-conditioning units and solar water heaters, and the insulation of water heaters. Copper is a good conductor of heat, making it ideal for transferring heat from the surrounding or sun to the fluid inside.

(2)  Other poor conductors include sawdust, glass wool, and styrofoam, which trap air and are used in refrigerators and ice boxes.

(3) Clothes made from wool, felt, and synthetic sponge are comfortable in winter due to their ability to trap air within their texture, preventing heat loss to colder surroundings. Two bedsheets combined to cover the body help retain body heat better than a single bedsheet of double the thickness, as the layer of air between the sheets reduces thermal conduction. A blanket coupled with a bedsheet is a cheaper alternative to using two blankets.

(4) The handle of a cooking utensil is made of a bad conductor of heat, such as ebonite, to protect the hand from hot utensils.

The opposition to the flow of heat through a body is called thermal resistance.

Definition: Thermal resistance is defined as the ratio of the temperature difference between the two parallel faces of a material to the rate at which heat is conducted through those surfaces. If Q is the quantity of heat conducted between two parallel surfaces of a slab in time t, the heat conduction rate is

\(\frac{Q}{t}\) = kA\((\frac{T_1-T_2}{x})\)

where k ≡ the thermal conductivity of the material,

A ≡ the cross-sectional area of each surface,

x ≡ the perpendicular distance between the two surfaces,

T₁ − T₂ ≡ the temperature difference between the two surfaces.

Then, the thermal resistance of the slab is,

RT = \(\frac{T_1-T_2}{Q/t}\)  = \(\frac{x}{kA}\)

SI unit: The kelvin second per joule k-s/J

Dimensions = [RT] = [M^-1L^-2T³K¹].

• Radiation is the process of heat transfer through electromagnetic waves without the need for any medium or heating.
• Thermal radiation, produced by body particles, is the radiation produced by a body above absolute zero of temperature.
• The spectrum of thermal radiation, which ranges from the far infrared to the extreme ultraviolet region, depends on the body's temperature.
• Infrared waves, with wavelengths ranging from 700 nm to 1 mm, are sensed as heat.
• When thermal radiations are incident on a body, they increase molecular motions, causing a rise in temperature.
• The transfer of heat by radiation is a two-fold process: the conversion of thermal energy into waves and the reconversion of waves into thermal energy by the body.
• Thermal radiation, as electromagnetic waves, has the same speed in free space as light, nearly 3 x 10⁸ m/s, making it the most rapid mode of heat transfer.

Newton's law of cooling: The time rate of fall of temperature of a body is directly

proportional to the excess temperature over the surroundings, provided the excess is small.

If T and T₀ are the temperatures of a hot body and the surroundings, respectively, then by Newton's law of cooling, the time rate of fall of temperature of the body, i.e., the time rate of cooling, is

dT/dt ∝ (T − T₀)

∴ dT/dt = C(T − T₀)

where C is a proportionality constant in a particular case.

Experimental verification:

• A calorimeter is filled up to two third of its capacity with boiling water and is covered.
• A thermometer is fixed through a hole in the lid and its position is adjusted so that the bulb of the thermometer is fully immersed in water.
• The calorimeter vessel is kept in a constant temperature enclosure or just in open air since room temperature will not change much during experiment.
• The temperature on the thermometer is noted at one minute interval until the temperature of water decreases by about 25 °C.

• A graph of temperature T (along y-axis) is plotted against time t (along x-axis). This graph is called cooling curve (Fig).
• From this graph we can infer how the cooling of hot water depends on the difference of its temperature from that of its surroundings.
• We will also notice that initially the rate of cooling is higher and it decreases as the temperature of the water falls.
• A tangent is drawn to the curve at suitable points on the curve. The slope of each tangent (dT/dt) gives the rate of fall of temperature at that temperature.
• Taking (0,0) as the origin, if a graph of dT/dt is plotted against corresponding temperature difference (T-T₀), the curve is a straight line as shown in Fig (a) and (b).

The above activity shows that a hot body loses heat to its surroundings in the form of heat radiation. The rate of loss of heat depends on the difference in the temperature of the body and its surroundings.

By either graph, we can conclude that dT/dt ∝ (T − T₀)

This verifies Newton's law of cooling.

Thermal stress is the stress induced in a body due to change in temperature. Thermal stress is set up in a body when it is subjected to a change in temperature but is restricted from expanding or contracting.

Disadvantages of thermal stress :

• Temperature differences cause uneven expansion or contraction, leading to thermal stresses. For example, hot water in a glass vessel can cause cracking due to uneven expansion.
• Insufficient thermal expansion joints can cause railway tracks and bridge spans to distorted in summer, while electrical transmission lines may snap in cold weather due to contraction.
• Long metal pipelines with loops should be given proper space for expansion to avoid significant thermal stress, which can damage the pipelines and cause accidents. Proper slack is essential for preventing these issues.

Wood, cellulose, wool, hemp and foam are commonly used as thermal insulators.

An ideal thermal insulator should have the lowest thermal conductivity and highest heat resistance. Thus, less insulating material will be required.

Given : V = 1 x 10^-4 m³, T₁ = 30°C, T₂ = 100°C,

γ_G = 1.2 x 10^-5 °C^-1, γ_L= 75 x 10^-5 °C^-1

Increase in the volume of the glass flask

= Vγ_G(T₂ − T₁)

Increase in the volume of the liquid

= Vγ_L(T₂ − T₁)

The volume of the liquid that will overflow

= Vγ_L(T₂ − T₁) − Vγ_G(T₂ − T₁)

= V(T­ − T₁) (γ_{L −} γ_G)

= (1 x 10^-4) (100 − 30) (75 − 1.2) x 10^-5

= 70 x 73.8 x 10^-9 = 5166 x 10^-9

= 5.166 x 10^-6 m³

Given : m_L = 2.0 kg, m_c = 4.0 kg, S_L =128 J Kg^-1 K^-1, S_C =387 J Kg^-1 k^-1, T_L =30 K, T_c = 5 K,

where T denotes the rise in temperature

Q_L = m_Ls_L T_L = 2 x 128 x 30 = 7680 J

Q_C = m_Cs_C T_C =4 x 387 x 5 =774O J

Copper block will require more energy.

Given : L_v = 2.26 X 10⁶ J/kg, m = 5.0 x 10^-3 kg

Q = mL_v

= (5.0 x 10^-3) (2.26 x 10⁶)

= 11.3 x 10³ J

This is the required energy.

Given : \((\frac{dT}{dt})_1\) = 0.05 °C/s, T₁ = 70 °C,

 \((\frac{dT}{dt})_2\) = 0.025 °C/_S, T₂ = 50 °C, T₃ = 40°C

(a) \(\frac{dT}{dt}\) = C(T − T₀)

\((\frac{(dT/dt)_1}{(dT/dt)_2}\) =  \(\frac{T_1-T_0}{T_2-T_0}\)

\(\frac{0.05}{0.025}=\frac{70-T_0}{50-T_0}\)

2 = \(\frac{70-T_0}{50-T_0}\)

100 – 2T₀ = 70 – T₀

T₀ = 100 – 70 = 30°C

This is the temperature of the surroundings.

(b) \((\frac{(dT/dt)_3}{(dT/dt)_2}\) = \(\frac{T_3-T_0}{T_2-T_0}\)

= \(\frac{40-30}{50-30}=\frac{1}{2}\)

∴ \((\frac{dT}{dt})_3\) = \(\frac{1}{2}(\frac{dT}{dt})_2\)   = \(\frac{1}{2}\) x 0.025 = 0.0125 °C

This is the rate of cooling when the temperature of the metal sphere is 40°C.

Given: p = constant

pV = nRT

for n moles of a gas. Here n = constant.

At constant pressure,

V → 0 as T → 0 K

The volume of the gas will be ideally zero at

0 K, i.e., — 273.15 °C.

Given: T₁ = 27°C, T₂ = 45 °C, L₁= 10 m, ∝ = 1.2 x 10^-5 K^-1 = 1.2 x 10^-5 °C^-1

Increase in the length of a rail section,

L₂ − L₁= ∝L₁ (T₂ − T₁)

= (1.2 x 10^-5) (10) (45 − 27)

= (1.2) (18) x 10^-4

= 21.6 x 10^-4 m

= 2.16 mm

This is the required gap between the consecutive rail sections.

Given : d₁ = 1.47 m, d₂ = 1.5 m, T₁ = 27°C, ∝ = 1.2 x 10^-5 K^-1 =1.2 x 10^-5 °C^-1

Circumference = d

d₂ – d₁ = ∝ d₁ (T₂ – T₁)

∴ T₂ – T₁ = \frac{d_2-d_1}{∝d_1}\) = \frac{1.5-1.47}{1.2×10^{-5}×1.47}\) = \(\frac{3000}{1.2×1.47}\) = 1701 °C

T₂ = T₁ + 1701 = 27 + 1701 = 1728°C

This is the required temperature.

Rise of 100 °C corresponds to 200 °X - 20 °X = 180 °X

A rise of 62 °C would correspond to a rise of

\(\frac{180}{100}\) x (62 °X) = 1.8 x 62 = 111.6 °X on the scale X.

The boiling point of the liquid

= 20 °X + 111.6 °X = 131.6 °X

Given: T₁ = 900°C = (900 + 273.15) K = 173.15 K, p₂ = p₁/2 , V₂ = V₁/2

pV = nRt

p₁V₁ = nRT₁ and p₂V₂ = nRT₂

\(\frac{T_2}{T_1}=\frac{p_2V_2}{p_1V_1}\) = \((\frac{p_2}{p_1})(\frac{V_2}{V_1})\) = \((\frac{1}{2})(\frac{1}{2})\) = \(\frac{1}{4}\) 

Given : Difference in lengths = 1.5 m at all temperatures, am = 24.5 X 10^—6/°C,

∝ = 11.9 x10^—6/°C

L_tI – L_0I = ∝_I L_0I(T – T₀)

L_tAl – L_0Al = ∝_Al L_0Al(T – T₀)

L_tI – L_0I = L_tAl – L_0Al  by the data

∴ ∝_I L_0I = ∝_Al L_0Al

∴ \(\frac{L_{0I}}{L_{0Al}}=\frac{∝_{Al}}{∝_I}\) = \(\frac{24.5×10^{-6}}{11.9×10^{-6}}\) = \(\frac{24.5}{11.9}\)

∴ L_0l = \(\frac{24.5}{11.9}\)L_{0Al                               (1)}               

Now, L_0I – L_0Al = 1.5m       (2)

∴ From (1) and (2), L_0Al = \((\frac{24.5}{11.9}-1)L_{oAl}\) = 1.5

∴ L_0Al = \(\frac{11.9×1.5}{24.5-11.9}=\frac{11.9×1.5}{12.6}\) = 1.4167m. 

This is the length of the aluminium rod at 0 °C.

L_0I = L_0Al + 1.5 = 1.4167 + 1.5 = 2.9167 m

This is the length of the iron rod at 0 °C.

Given : Q = 50 cal, m = 6 kg, T₁ = 20 °C, T₂ = 62 °C

Q = ms(T₂ - T₁)

s = \(\frac{Q}{m(T_2-T_1)}\)

= \(\frac{50}{6(62-20)}\)

= 0.1984 cal/kg⁰C

This is the specific heat of the metal.

Given : T₁ − T₂ = 30 °C, Q/t = 1500 cal/s

The thermal resistance of the copper rod,

R_T = \(\frac{T_1-T_2}{Q/t}\) = \(\frac{30}{1500}\) = 0.02 \(\frac{°C}{cal/s}\)

Given: t₁ = 20 minutes, t₂ = 90 minutes, s = 1 cal/g.°C

Q₁ = ms (t₂ – t₁)

[Heating water from 0 °C to 100°C]

= m x 1(100°C - 0°C) = 100 m cal/g

Q₂ = mL_v

[Converting water at 100 °C into steam]

Now, Q₂ = \(\frac{Q_1}{20min}×90min\) 

As the source supplies the energy at a constant rate,

\(\frac{Q_2}{90min}= \frac{Q_1}{20min}\)

Q₂ = \(\frac{9}{2}\)Q₁ = 4.5 Q₁

mL_v = 4.5 x 100 m

L_v = 450 cal/g

This is the latent heat of the vaporization water.

Given: \(\frac{Q}{At}\) = 400 kcal/minute.m² = 400/60 kcal/s.m² = 20/3 kcal/s-m², x = 4 x 10^-2 m, k = 0.026 kcal/m.s.K

k = \((\frac{Q}{At})(\frac{x}{T_1-T_2})\)

T₁ – T₂ = \((\frac{Q}{At})(\frac{x}{k})\) = \((\frac{20}{3})(\frac{4×10^{-2}}{0.026})\) = 10.26 K

This is the required temperature difference.

Given : \((\frac{dT}{dt})_1=\frac{80-60}{6}=\frac{10}{3}\frac{°C}{min}\)

T₁ = \(\frac{80+60}{2}\) = 70°C, T₀ = 30°C

In part 2, T₂ = \(\frac{60+40}{2}\) = 50 °C

and (dT)₂ = 60°C − 40°C = 20°C

\(\frac{dT}{dt}\) = C (T₁ – T₀)

\(\frac{(dT/dt)_2}{(dT/dt)_1}\) = \(\frac{T_2-T_0}{T_1-T_0}\) = \(\frac{50-30}{70-30}\) = 1/2 

∴ \((\frac{dT}{dt})_2\) = \((\frac{dT}{dt})_2 ×\frac{1}{2}\) = \(\frac{10}{3}×\frac{1}{2}\) = 5/3

∴ \(\frac{dT_2}{dt_2}=\frac{5}{3}\)

∴ dt₂ = \(\frac{3}{5}\) x dT₂ = \frac{3}{5}\) x 20

∴ dt₂ = 12 minutes

This is the required time.