NEET Physics Solutions For Class 11 Chapter 1 Calorimetry And Thermal Expansion

NEET Physics Class 11 Chapter 1 Calorimetry And Thermal Expansion Heat

The energy that is being transferred between two bodies or between adjacent parts of a body as a result of temperature difference is called heat. Thus, heat is a form of energy.

• It is energy in transit whenever temperature differences exist. Once it is transferred, it becomes the internal energy of the receiving body.
• It should be clearly understood that the word “heat” is meaningful only as long as the energy is being transferred.
• Thus, expressions like “heat in a body” or “heat of a body” are meaningless. Heat transfer from a body at high temperature to low temperature.

When we say that a body is heated it means that its molecules begin to move with greater kinetic energy. So, it is the energy of molecular motions.

S.Ι. unit of heat energy is joule (J). Another practical unit of heat energy is calorie (cal).

1 calorie = 4.18 joules.

1 calorie: The amount of heat needed to increase the temperature of 1 gm of water from 14.5 to 15.5 ºC at one atmospheric pressure is 1 calorie.

Mechanical Equivalent of Heat

In the early days, the heat was not recognized as a form of energy. The heat was supposed to be something needed to raise the temperature of a body or to change its phase. The calorie was defined as the unit of heat.

• A number of experiments were performed to show that the temperature may also be increased by doing mechanical work on the system.
• These experiments established that heat is equivalent to mechanical energy and measured how much mechanical energy is equivalent to a calorie. If mechanical work W produces the same temperature change as heat H, we write,

W = JH

where J is called the mechanical equivalent of heat. J is expressed in joule/calorie. The value of J gives how many joules of mechanical work is needed to raise the temperature of 1 g of water by 1°C. It is a conversion factor and not a physical quantity.

Question 1. What is the change in potential energy (in calories) of a 10 kg mass when it falls through 10 m?

Solution: Change in potential energy

Mass = 10 kg

Gravitation = 10

Height = 10 m

ΔU = mgh = 10 × 10 × 10 = 1000 J

⇒ \(\frac{1000}{4.186} \text { cal }\)

NEET Physics Class 11 Chapter 1 Specific Heat

The specific heat of a substance is equal to the heat gained or released by that substance to raise or fall its temperature by 1ºC for a unit mass of a substance.

When a body is heated, it gains heat. On the other hand, heat is lost when the body is cooled. The gain or loss of heat is directly proportional to:

1. The mass of the body ΔQ ∝ m
2. Rise or fall of temperature of the body ΔQ ∝ Δ T

ΔQ ∝ m Δ T or ΔQ = m s Δ T

or dQ = m s d T or Q = m ∫s d T.

where s is a constant and is known as the specific heat of the body s = \(\frac{Q}{m \Delta T}\) joule/kg-kelvin and C.G.S. unit is cal./gm °C.

Qm TΔ. S.Ι. unit of s is

Specific heat of water : S = 4200 J/kgºC = 1000 cal/kgºC = 1 Kcal/kgºC = 1 cal/gmºC

Specific heat of steam = half of specific heat of water = specific heat of ice

Question 1. Calculate the heat required to increase the temperate of 1 kg water by 20ºC

Solution :

Heat required = ΔQ = msΔθ

∵ S = 1 cal/gmºC = 1 Kcal/kgºC

= 1 × 20 = 20 Kcal.

Heat capacity or Thermal capacity: The heat capacity of a body is defined as the amount of heat required to raise the temperature of that body by 1°. If ‘m’ is the mass and ‘s’ the specific heat of the body, then

Heat capacity = m s.

Units of heat capacity in the CGS system is, cal ºC^-1; the SI unit is, JK^-1

Important Points:

1. We know, s = \(\frac{Q}{m \Delta T}\), if the substance undergoes the change of state which occurs at constant temperature (Isothermal ΔT = 0), then s = Q/0 = ∞. Thus the specific heat of a substance when it melts or boils at constant temperature is infinite.
2. If the temperature of the substance changes without the transfer of heat adiabatic (Q = 0) then s = \(\frac{Q}{m \Delta T}\) = 0. Thus when liquid in the thermos flask is shaken, its temperature increases without the transfer of heat, and hence the specific heat of liquid in the thermos flask is zero.
3. To raise the temperature of saturated water vapors, heat (Q) is withdrawn. Hence, the specific heat of saturated water vapors is negative. (This is for your information only and not in the course)
4. The slight variation of specific heat of water with temperature is shown in the graph at 1-atmosphere pressure. Its variation is less than 1% over the interval from 0 to 100ºC.

Relation between Specific heat and Water equivalent: It is the amount of water that requires the same amount of heat for the same temperature rise as that of the object

ms ΔT = m_W S_W ΔT ⇒ mW = \(\frac{\mathrm{ms}}{\mathrm{s}_{\mathrm{W}}}\)

In calorie s_W = 1

∴ m_W = ms

m_W is also represented by W

So, W = ms.

Phase change: Heat required for the change of phase or state,

Q = mL, L = latent heat.

Latent heat (L): The heat supplied to a substance that changes its state at constant temperature is called latent heat of the body.

Latent heat of Fusion (L_f ): The heat supplied to a substance which changes it from solid to liquid state at its melting point and 1 atm. pressure is called latent heat of fusion. The latent heat of the fusion of ice is 80 kcal/kg

Latent heat of vaporization (L_v): The heat supplied to a substance that changes it from liquid to vapor state at its boiling point and 1 atm. pressure is called latent heat of vaporization. The latent heat of vaporization of water is 540 kcal kg^-1.

If in question latent heat of ice or steam is not mentioned and to solve the problem, it is required to assume them, we should consider the following values.

Latent heat of ice: L = 80 cal/gm = 80 Kcal/kg = 4200 × 80 J/kg

Latent heat of steam: L = 540 cal/gm = 540 Kcal/kg = 4200 × 540 J/kg

The given behavior of a solid substance when it is continuously heated is shown. Various parts show.

OA − solid state, AB − solid + liquid state (Phase change)

BC − liquid state, CD − liquid + vapor state (Phase change)

DE − vapor state

When there is no change of state for example., OA, BC, DE

Slope, \(\frac{\Delta T}{\Delta \mathrm{Q}}=\frac{1}{\mathrm{~ms}} \quad \Rightarrow \quad \frac{\Delta \mathrm{T}}{\Delta \mathrm{Q}} \propto \frac{1}{\mathrm{~s}}\)

The mass (m) of a substance is constant. So, the slope of T – Q graph is inversely proportional to specific heat. In the given diagram.

(slope) OA > (slope) DE

then (s)_OA < (s)_DE

when there is a change of state for example., AB and CD

ΔQ = mL

If (length of AB) > (length of CD)

then (latent heat of AB) > (latent heat of CD)

Question 2. Find the amount of heat released when 1 kg steam at 200ºC is converted into –20ºC ice.

Answer:

Heat required ΔQ = heat release to convert steam at 200 ºC into 100ºC steam + heat release to convert 100ºC steam into 100ºC water + heat release to convert 100º water into 0ºC water + heat release to convert 0 ºC water into 0ºC ice + 0ºC ice converted to – 20ºC ice.

⇒ \(\Delta Q=1 \times \frac{1}{2} \times 100+540 \times 1+1 \times 1 \times 100+1 \times 80+1 \times \frac{1}{2}=780(\mathrm{Kcal})\)

NEET Physics Class 11 Chapter 1 Superficial Or Areal Expansion

When a solid is heated and its area increases, then the thermal expansion is called superficial or areal expansion. Consider a solid plate of area A0. When it is heated, the change in the area of the plate is directly proportional to the original area A₀ and the change in temperature ΔT.

dA = βA₀dT or ΔA = β A₀ Δ T; β is called real expansion

β = \(\frac{\Delta \mathrm{A}}{\mathrm{A}_0 \Delta \mathrm{T}}\) Unit of β is ºC^-1 or K^-1.

A = A₀(1 + β Δ T)

Where A is an area of the plate after heating.

it follows that β = 2α.

Question 1. A plane lamina has an area of 2m² at 10ºC then what is its area at 110ºC Its superficial expansion is 2 × 10^-5/C

Answer:

Area of plane lamina = 2m²

Superficial expansion = 2 × 10^-5/C

A = A₀ ( 1 + β Δ θ ) = 2 {1 + 2 × 10⁵ × (110 – 10)}

= 2 × {1 + 2 × 10^-3} m²

NEET Physics Class 11 Chapter 1 Volume Or Cubical Expansion

When a solid is heated and its volume increases, then the expansion is called volume expansion or cubical expansion. Let us consider a solid or liquid whose original volume is V₀. When it is heated to a new volume, then the change ΔV

dV = γV₀dT or ΔV = γ V₀ Δ T

⇒ \(\gamma=\frac{\Delta V}{V_0 \Delta T}\) Unit of γ is ºC^-1 or K^-1.

V = V₀(1 + γ Δ T)

where V is the volume of the body after heating.

It can be shown easily that γ = 3α for isotropic solids.

Question 1. The volume of the glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature? The coefficient of cubical expansion of mercury and glass is 1.8 × 10^-4/°C and 9.0 × 10^-6/°C respectively.

Answer:

Let the volume of the glass vessel at 20ºC be Vgand volume of mercury at 20ºC be Vm so the volume of remaining space is = V_g– V_m

It is given constant so that

V_g– V_m= V_g’ – V’_m

where V₀‘ and V_m‘ are final volumes.

V_g– V_m= V_g{1 + γ_g Δθ} – V_m{1 + γ_Hg Δθ} ⇒ V_gγ_g= V_mγ_Hg

⇒ V_m = \(V_m=\frac{100 \times 9 \times 10^{-6}}{1.8 \times 10^{-4}} V_m=50 \mathrm{cc}\)

Relation Between α, β And γ

1. For isotropic solids: α : β : γ = 1 : 2 : 3 or \(\frac{\alpha}{1}=\frac{\beta}{2}=\frac{\gamma}{3}\)
2. For non-isotropic solid β = α₁+ α₂ and γ = α₁+ α₂+ α₃. Here α₁, α_2, and α₃ are coefficients of linear expansion in the X, Y, and Z direction.

Question 1. If the percentage change in length is 1% with the change in temperature of a cuboid object (l × 2l × 3l) then what is the percentage change in its area and volume?

Answer:

Percentage change in length with change in temperature = %l

⇒ \(\frac{\Delta \ell}{\ell} \times 100=\alpha \Delta \theta \times 100=1\)

change in area ⇒ % A = \(\frac{\Delta \mathrm{A}}{\mathrm{A}} \times 100=\beta \Delta \theta \times 100\)

⇒ 2 (α Δ θ × 100) % A = 2%

Change in volume

% V = \(\) × 100 = V Δ θ × 100 = 3 (α Δ θ × 100) % V = 3 %

NEET Physics Class 11 Chapter 1 Variation Of Density With Temperature

As we know mass = volume × density.

The mass of a substance does not change with a change in temperature so with an increase in temperature, volume increases so density decreases and vice-versa.

d = \(\frac{d_0}{(1+\gamma \Delta T)}\)

For solids values of γ are generally small so we can write d = d₀(1 − γ ΔT) (using binomial expansion).

Note:

1. γ for liquids is of the order of 10−3.
2. Anomalous expansion of water:

For water density increases from 0 ºC to 4 ºC so γ is negative and for 4 ºC to higher temperature γ is positive. At 4 ºC density is maximum. This anomalous behavior of water is due to the presence of three types of molecules i.e. H₂O, (H₂O)₂and (H₂O)₃ having different volumes/masses at different temperatures.

This anomalous behavior of water causes ice to form first at the surface of a lake in cold weather. As winter approaches, the water temperature decreases initially at the surface. The water there sinks because of its increased density.

Consequently, the surface reaches 00C first and the lake becomes covered with ice. Aquatic life is able to survive the cold winter as the lake bottom remains unfrozen at a temperature of about 4°C.

Question 1. The densities of wood and benzene at 0°C are 880 kg/m³ and 900 kg/m³ respectively. The coefficients of volume expansion are 1.2 × 10^-3/°C for wood and 1.5 × 10^-3/°C for benzene. At what temperature will a piece of wood just sink in benzene?

Answer:

At just sink gravitation force = up thrust force

⇒ mg = F_B

⇒ Vp₁g = Vp₂g

⇒ p₁= p₂

⇒ \(\frac{880}{1+1.2 \times 10^{-3} \theta}=\frac{900}{1+1.5 \times 10^{-3} \theta}\)

⇒ θ = 83º C

NEET Physics Class 11 Chapter 1 Apparent Expansion Of A Liquid In A Container

Initially container was full. When temperature increases by ΔT,

volume of liquid V_L= V₀(1 + γ_L Δ T)

volume of container V_C= V₀(1 + γ_C Δ T)

So overflow volume of liquid relative to the container

ΔV = V_L− V_C ΔV = V₀(γ_L− γ_C) ΔT

So, the coefficient of apparent expansion of liquid w.r.t. Container

γ_apparent = γ_L− γ_C.

In case of expansion of liquid + container system:

if γ_L> γ_C ⎯→ level of liquid rise

if γ_L< γ_C ⎯→ level of liquid fall

Increase in height of liquid level in a tube when the bulb was initially completely filled

h = \(h=\frac{\text { volume of liquid }}{\text { area of tube }}=\frac{V_0\left(1+\gamma_L \Delta T\right)}{A_0\left(1+2 \alpha_S \Delta T\right)}=h_0\left\{1+\left(\gamma_L-2 \alpha_S\right) \Delta T\right\}\)

h = h₀{ 1 + ( γ_L– 2α_S) ΔT}

where h₀= original height of liquid in a container

α_S= linear coefficient of expansion of container.

Question 1. A glass vessel of volume 100 cm³ is filled with mercury and is heated from 25°C to 75°C. What volume of mercury will overflow? The coefficient of linear expansion of glass = 1.8 × 10^-6/°C and the coefficient of volume expansion of mercury is 1.8 × 10^-4/°C.

Answer:

ΔV = V₀(γ_L– γ_C) ΔT = 100 × {1.8 × 10^-4 – 3 × 1.8 × 10^-6} × 50

ΔV = 0.87 cm³

Variation Of Force Of Buoyancy With Temperature

If the body is submerged completely inside the liquid

For solid, Buoyancy force F_B= V₀ dL g

V₀= Volume of the solid inside the liquid,

d_L= density of liquid

Volume of body after increase its temperature V = V₀[1 + γ_S Δθ],

Density of body after increase its temperature d′_L = \(\frac{d_{\mathrm{L}}}{\left[1+\gamma_{\mathrm{L}} \Delta \theta\right]}\)

Buoyancy force of body after increasing its temperature, \(F_B^{\prime}=V d_L^{\prime} g \quad \Rightarrow \frac{F_B^{\prime}}{F_B}=\frac{\left[1+\gamma_S \Delta \theta\right]}{\left[1+\gamma_L \Delta \theta\right]}\)

If γ_S< γ_L then F′_B< F_B

(Buoyant force decreases) or apparent weight of the body in liquid gets increased

[W − F′_B> W − F_B].

Question 1. A body is floating on the liquid if we increase temperature then what changes occur in buoyancy force? (Assume the body is always in floating condition)

Answer:

The body is in equilibrium

So, mg = B (Boyant Force)

The gravitational force does not change with a change in temperature. So buoyancy force remains constant.

By increasing temperature density of the liquid decreases so the volume of the body inside the liquid increases to keep the Buoyance force constant and equal to the gravitational force)

Question 1. In the previous question discuss the case when the body move downward, upwards and remains at same position when we increases temperature.

Answer:

Let f = fraction of volume of body submerged in liquid

⇒ \(\mathrm{f}=\frac{\text { volume of body submerged in liquid }}{\text { total volume of body }}\)

⇒ \(f_1=\frac{v_1}{v_0} \text { at } \theta_1{ }^{\circ} \mathrm{C}\)

⇒ \(f_2=\frac{v_2}{v_0\left(1+3 \alpha_S \Delta \theta\right)}\) at θ2ºC

for equilibrium mg = B = v1d1g = v2d2g. d

so v₂ = \(\frac{v_1 d_1}{d_2}\)

⇒ \(d_2=\frac{d_1}{1+\gamma_L \Delta \theta}=v_1\left(1+\gamma_L \Delta \theta\right)\)

⇒ \(f_2=\frac{v_1\left(1+\gamma_L \Delta \theta\right)}{v_0\left(1+3 \alpha_s \Delta \theta\right)}\)

where Δθ = θ₂– θ₁

Case 1: Body move downward if f₂> f₁

means γ_L> 3α_S

Case 2: Body moves upwards if f₂< f₁

means γ_L< 3α_S

Case 3: The body remains in the same position

if f₂= f₁

means γ_L= 3α_S

Bimetallic Strip

Two strips of different metals are welded together to form a bimetallic strip, when heated uniformly it bends in the form of an arc, and the metal with a greater coefficient of linear expansion lies on the convex side. The radius of the arc is thus formed by bimetal:

⇒ \(\ell_0\left(1+\alpha_1 \Delta \theta\right)=\left(\mathrm{R}-\frac{\mathrm{d}}{2}\right) \theta\)

⇒ \(\ell_0\left(1+\alpha_2 \Delta \theta\right)=\left(R+\frac{d}{2}\right) \theta\)

⇒ \(\frac{1+\alpha_2 \Delta \theta}{1+\alpha_1 \Delta \theta}=\frac{R+\frac{d}{2}}{R-\frac{d}{2}}\)

⇒ \(R=\frac{d}{\left(\alpha_2-\alpha_1\right) \Delta \theta}\)

⇒ \(\Delta \theta=\text { change in temperature }=\theta_2-\theta_1\)

• A bimetallic strip, consisting of a strip of brass and a strip of steel welded together as shown in figure (3). At temperature T₀ in figure and figure  (2). The strip bends as shown at temperatures above the reference temperature.
• Below the reference temperature, the strip bends the other way. Many thermostats operate on this principle, making and breaking an electrical constant as the temperature rises and falls as shown in Figure (2).

Applications Of Thermal Expansion

1. A small gap is left between two iron rails of the railway.
2. Iron rings are slipped on the wooden wheels by heating the iron rings
3. The stopper of a glass bottle jammed in its neck can be taken out by heating the neck.
4. The pendulum of a clock is made of invar [an alloy of zinc and copper].

Temperature

Temperature may be defined as the degree of hotness or coldness of a body. Heat energy flows from a body at a higher temperature to that at a lower temperature until their temperatures become equal. At this stage, the bodies are said to be in thermal equilibrium.

Measurement of Temperature: The branch of thermodynamics which deals with the measurement of temperature is called thermometry.

• A thermometer is a device used to measure the temperature of a body.
• The substances like liquids and gases that are used in the thermometer are called thermometric substances. Suppose a physical quantity X varies linearly with temperature then

⇒ \(X_t=X_0(1+\alpha \mathrm{t}) \Rightarrow \quad t=\frac{X_t-X_0}{X_0 \alpha}\)

α is constant and X0 is the value of X at the reference temperature set at 0º. In the case of absolute scale

⇒ \(\frac{X_T}{X_0}=\frac{T}{T_0} \quad \Rightarrow \quad T=T_0 \frac{X_T}{X_0}=273.16 \frac{X_T}{X_0} K\)

Different Scales of Temperature: A thermometer can be graduated into the following scales.

1. The Centigrade or Celsius scale (ºC)
2. The Fahrenheit scale (ºF)
3. The Reaumer scale (ºR)
4. Kelvin scale of temperature (K)

Comparison Between Different Temperature Scales:

The formula for the conversion between different temperature scales is:

⇒ \(\frac{\mathrm{K}-273}{100}=\frac{\mathrm{C}}{100}=\frac{\mathrm{F}-32}{180}=\frac{\mathrm{R}}{80}\)

A general formula for the conversion of temperature from one scale to another:

⇒ \(=\frac{\text { Temp. on other scale }\left(\mathrm{S}_1\right) \text {-Lower fixed point }\left(\mathrm{S}_1\right)}{\text { Upper fixed point }\left(\mathrm{S}_2\right) \text {-Lower fixed point }\left(\mathrm{S}_1\right)}\)

⇒ \(=\frac{\text { Temp. on other scale }\left(\mathrm{S}_2\right) \text {-Lower fixed point }\left(\mathrm{S}_2\right)}{\text { Upper fixed point }\left(\mathrm{S}_2\right) \text {-Lower fixed point }\left(\mathrm{S}_2\right)}\)

Thermometers: Thermometers are devices that are used to measure temperatures. All thermometers are based on the principle that some physical property of a system changes as the system temperature changes.

Required properties of good thermometric substance.

1. Non-sticky (absence of adhesive force)
2. Low melting point (in comparison with room temperature)
3. High boiling temperature
4. The coefficient of volumetric expansion should be high (to increase accuracy in measurement).
5. Heat capacity should be low.
6. Conductivity should be high Mercury (Hg) suitably exhibits the above properties.

Types Of Thermometers

The Constant-volume Gas Thermometer: The standard thermometer, against which all other thermometers are calibrated, is based on the pressure of a gas in a fixed volume.

The figure shows such a constant volume gas thermometer; it consists of a gas-filled bulb connected by a tube to a mercury manometer.

T = \((273.16 \mathrm{~K})\left(\lim _{{gas} \rightarrow 0} \frac{p}{p_3}\right)\)

P = Pressure at the temperature being measured,

P₃ = pressure when the bulb is in a triple point cell.

Question 1. The readings of a thermometer at 0ºC and 100ºC are 50 cm and 75 cm of mercury column respectively. Find the temperature at which its reading is 80 cm of mercury column.

Answer:

By using formula

⇒ \(\frac{80-50}{75-50}=\frac{T-0}{100-0}\)

⇒ T = 120ºC

Question 2. A bullet of mass 10 gm in moving with speed 400m/s. Find its kinetic energy in calories.

Answer:

⇒ \(\Delta k=\frac{1}{2} \times \frac{10}{1000} \times 400 \times 400=800\)

⇒ \(\frac{800}{4.2}=\mathrm{Cal}\) = 191.11 Cal.

Question 3. Calculate the amount of heat required to convert 1 kg steam from 100ºC to 200ºC steam.

Answer:

Heat req. = 1 × \(\frac{1}{2}\)× 100 = 50 kcal

Question 4. Calculate the heat required to raise the temperature of 1 g of water through 1ºC.

Answer:

Heat req. = 1 × 10^-3 × 1 × 1 = 1 × 10^-3 kcal

Question 5. 420 J of energy supplied to 10 g of water will raise its temperature by

Answer:

⇒ \(\frac{420 \times 10^{-3}}{4.20}=10 \times 10^{-3} \times 1 \times \Delta \mathrm{t}=10^{\circ} \mathrm{C}\)

Question 6. The ratio of the densities of the two bodies is 3: 4 and the ratio of specific heats is 4 : 3 . Find the ratio of their thermal capacities for unit volume.

Answer:

⇒ \(\frac{\rho_1}{\rho_2}=\frac{3}{4}, \frac{s_1}{s_2}=\frac{4}{3}\)

⇒ \(\theta=\frac{\mathrm{m} \times \mathrm{s}}{\mathrm{m} / \mathrm{\rho}}\)

⇒ \(\frac{\theta_1}{\theta_2}=\frac{s_1}{s_2} \times \frac{\rho_1}{\rho_2}=1: 1\)

Question 7. Heat is released by 1 kg steam at 150ºC if it converts into 1 kg water at 50ºC.

Answer:

H = 1 ×\(\frac{1}{2}\)× 50 + 1 × 540 + 1 × 1 × 50

= 540+75 = 615

Heat release = 615 Kcal.

Question 8. 200 gm water is filled in calorimetry of negligible heat capacity. It is heated till its temperature is increased by 20ºC. Find the heat supplied to the water.

Answer:

H = 200 × 10^-3 × 1 × 20 = 4 Kcal.

Heat supplied = 4000 cal = 4 Kcal

Question 9. A bullet of mass 5 gm is moving with a speed of 400 m/s. strike a target and energy. Then calculate the rise of the temperature of the bullet. Assuming all the loss in kinetic energy is converted into heat energy of the bullet if its specific heat is. 500J/kgºC.

Answer:

Kinetic energy = 12× 5 × 10^-3 × 400 × 400 = 5 × 10^-3 × 500 × ΔT

ΔT = 160º C

The rise in temperature is 160 ºC

Question 10. 1 kg ice at –10ºC is mixed with 1 kg water at 100ºC. Then find the equilibrium temperature and mixture content.

Answer:

Heat gained by 1 kg ice at – 10ºC to convert into 0ºC ice = 1 × \(\frac{1}{2}\)x10 = 5Kcal = 5000 cal

In the thermal equilibrium

5 + 1 × 80 + 1 × T = 1 × (100 – T)

85 = 100 – 2T ⇒ 2T = 15

⇒ \(\theta=\frac{15}{2}\) = 7.5 ºC, water, entire ice melts.

Question 11. 1 kg ice at –10º is mixed with 1kg water at 50ºC. Then find the equilibrium temperature and mixture content.

Answer:

The heat required by ice at –10ºC to convert it into 0ºC water

1 × \(\frac{1}{2}\)× 10 + 1 × 50 = 55 Kcal

The heat released by 1 kg of water to reduce its temperature from

50ºC to 0ºC = 1 × 1 × 50 = 50 kcal

Heat required > Heat released so, ice will not completely melt. Let m g ice melt then for equilibrium.

1 × \(\frac{1}{2}\) × 10 + 80 m = 50 m ⇒ 80 m = 45 ⇒ m = \(\frac{45}{80}\)

⇒ \(\text { Content of mixture }\left\{\begin{array}{ll}
\text { water } & \left(1+\frac{45}{80}\right) \mathrm{kg} \\
\text { ice } & \left(1-\frac{45}{80}\right) \mathrm{kg}
\end{array}\right\} \text { and temperature is } 0^{\circ} \mathrm{C}\)

Question 12. A small ring having a small gap is shown in the figure on heating what will happen to the size of the gap?

Answer:

The gap will also increase. The reason is the same as in the above example.

Question 13. 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.

Answer:

l1= 10(1 + 1 × 10^-5× 35) = 10.0035 m

Question 14. A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10^-5/°C.
Δ αΔt

Answer:

⇒ \(\frac{\Delta \ell}{\ell}=\frac{\ell_0 \alpha \Delta t}{\ell_0}\)= – 3.6 × 10^-4

Question 15. If the rod is initially compressed by Δl length then what is the strain on the rod when the temperature

1. Is increased by Δθ
2. Is decreased by Δθ.

Answer:

1. Strain = \(\frac{\Delta \ell}{\ell}+\alpha \Delta \theta\)
2. Strain = \(\left|\frac{\Delta \ell}{\ell}-\alpha \Delta \theta\right|\)

Question 16. A pendulum clock with having copper rod keeps the correct time at 20°C. It gains 15 seconds per day if cooled to 0°C. Calculate the coefficient of linear expansion of copper.

Answer:

⇒ \(\frac{15}{24 \times 60 \times 60}=\frac{1}{2} \alpha \times 20\)

⇒ \(\alpha=\frac{1}{16 \times 3600}=1.7 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

Question 17. A meter scale made of steel is calibrated at 20°C to give the correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. The coefficient of linear expansion of steel is 1.1 × 10^-5/°C.

Answer:

lt = 1 (1 – 1.1 × 10^-5 × 10) = 0.99989 cm

Question 18. A uniform solid brass sphere rotates with an angular speed ω₀ about a diameter. If its temperature is now increased by 100ºC. What will be its new angular speed? (Given α_B = 2.0 × 10^-5 perºC)

1. \(\frac{\omega_0}{1-0.002}\)
2. \(\frac{\omega_0}{1+0.002}\)
3. \(\frac{\omega_0}{1+0.004}\)
4. \(\frac{\omega_0}{1-0.004}\)

Answer:

⇒ \(\mathrm{I}_0 \omega_0=\mathrm{I}_{\mathrm{t}} \omega \mathrm{t}\)

⇒ \(M r_0{ }^2 \omega_0={Mr}_0{ }^2(1+2 \alpha \Delta \mathrm{T}) \omega_{\mathrm{t}}\)

⇒ \(\omega_t=\frac{\omega_0}{1+0.004}\)

Question 19. The volume occupied by a thin-wall brass vessel and the volume of a solid brass sphere is the same and equal to 1,000 cm³ at 0ºC. How much will the volume of the vessel and that of the sphere change upon heating to 20ºC? The coefficient of linear expansion of brass is α = 1.9 × 10^-5.

Answer:

V = V₀(1 + 3α ΔT) = 1.14 cm³ ⇒ 1.14 cm³ for both

Question 20. A thin copper wire of length L increases in length by 1%, when heated from temperature T₁ to T₂. What is the percentage change in area when a thin copper plate having dimensions 2L × L is heated from T₁ to T₂?

1. 1%
2. 3%
3. 4%
4. 2%

Answer: \(\ell_{\mathrm{f}}=\mathrm{L}(1+\alpha \Delta \mathrm{t})=\frac{\mathrm{L}_{\mathrm{f}}}{\mathrm{L}} \times 100=(1+\alpha \Delta t) \times 100=1 \%\)

⇒ \(A=2 L \times L(1+2 \alpha \Delta t)=\frac{A_f}{2 L \times L} \times 100=(1+2 \alpha \Delta t) \times 100=2 \%\)

Question 21. The density of water at 0°C is 0.998 g/cm³ and at 4°C is 1.000 g/cm³. Calculate the average coefficient of volume expansion of water in the temperature range of 0 to 4°C.

Answer: \(d_t=\frac{d_0}{1+\gamma \Delta t}\)

⇒ \(1=\frac{0.998}{1+\gamma \times 4}\)

= – 5 × 10^-4/°C

Question 22. A glass vessel measures exactly 10 cm × 10 cm × 10 cm at 0°C. it is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1.6 cm³ of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5 × 10^-6/°C

Answer:

ΔV = V_Hg – V_V

1.6 = 10³(γ_L× 10 – 10³ × 3 × 6.5 × 10^-6 × 10

γ_L = (1.6 + 0.195) × 10^-4 = 1.795 × 10^-4/°C

Question 23. A metal ball immersed in alcohol weighs W₁ at 00C and W₂ at 500C. The coefficient of cubical expansion of the metal is less than alcohol. Assuming that the density of the metal is large compared to that of the alcohol, find which of W₁ and W₂ is greater.

Answer:

⇒ \(\gamma_{\mathrm{M}}<\gamma_{\ell} \text { so, } \frac{F_{\mathrm{B}}^{\prime}}{F_{\mathrm{B}}}=\frac{\left[1+\gamma_{\mathrm{S}} \Delta \theta\right]}{\left[1+\gamma_{\ell} \Delta \theta\right]} F_{\mathrm{B}}^{\prime}<F_{\mathrm{B}}\)

So Apprent weight increased so, W₂> W₁

Question 24. In the figure strip brass or steel have a higher coefficient of linear expansion.

Answer:

Brass Strip; a strip of higher α is on the convex side.

Question 25. The upper and lower fixed points of a faulty thermometer are 5ºC and 105º C. If the thermometer reads 25ºC, what is the actual temperature?

Answer:

⇒ \(\frac{25-5}{100}=\frac{C-0}{100}\)

C = 20ºC

Question 26. At what temperature is the Fahrenheit scale reading equal to twice Celsius?

Answer:

⇒ \(\frac{F-32}{180}=\frac{C-0}{100}=\)

⇒ \(\frac{C-0}{100}\)

1 × -160 = 9x

x = 160º

Question 27. The temperature of a patient is 40º C. Find the temperature on a Fahrenheit scale.

Answer:

⇒ \(\frac{F-32}{180}\)

⇒ \(\frac{40-0}{100}\)

F = 104ºC