Class 11 - Physics - Thermal Properties of Matter

Question1.

The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively.

Express these temperatures on the Celsius and Fahrenheit scales.

Answer:

Celsius and Fahrenheit scales are related as

TF = (9/5) TC + 32    ... (ii)

For neon:

Tk= 24.57 K

TC= 24.57 – 273.15 = –248.58°C

TF = (9/5) TC + 32

= ((9/5) (-248.58) +32)

=415.44

For carbon dioxide:

Tk= 216.55 K

Tc= (216.55 – 273.15) = –56.60°C

TF = (9/5) TC + 32

= ((9/5) (-56.60) +32)

= -69.88

Question2.

Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?

Answer:

Triple point of water on absolute scale A, T1 = 200 A

Triple point of water on absolute scale B, T2 = 350 B

Triple point of water on Kelvin scale, Tk= 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A

T1 = Tk

200 A = 273.15 K

Therefore,

A = (273.15)/ (200)

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B

T2 = Tk

350 B = 273.15

B = (273.15)/ (350)

TA is triple point of water on scale A. TB is triple point of water on scale B

Therefore,

(273.15)/ (200) x TA = (273.15)/ (350) x TB

TA = (200)/ (350) TB

Therefore, the ratio TA : TB is given as 4: 7.

Comparison of Kelvin, Celsius and Fahrenheit temperature Scales Question 3.

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

R = Ro [1 + α (T – To)]

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K).

What is the temperature when the resistance is 123.4 Ω?

Answer:

Given:

R = Ro [1 + α (T – To)]   … (i)

Where Ro = initial resistance, R = final resistance, To = initial temperature and

T = final temperature; α = constant

At the triple point of water, T0 = 273.15 K

Resistance of lead, R0 = 101.6 Ω

At normal melting point of lead, T = 600.5 K

Resistance of lead, R = 165.5 Ω

Substituting these values in equation (i), we get:

R = Ro [1 + α (T – To)]

165.5 = 101.6[1 + α (600.5 - 273.15)]

1.629 = (1 + α (327.35))

Therefore, α = (0.629)/ (327.35)

=1.92 x 10-3 K-1

For resistance, R1 = 123.4 Ω

R1 = Ro [1 + α (T – To)]    where T=temperature when the resistance of lead is 123.4 Ω

123.4 = 101.6[1 + (1.92 x 10-3) (T – 273.15)]

1.214 = (1 + (1.92 x 10-3) (T – 273.15))

(0.214)/ (1.92 x 10-3) = (T – 273.15)

Therefore T = 384.61K

Question 4.

Answer the following:

(a) The triple-point of water is a standard fixed point in modern thermometry.

Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively.

On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K.

What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T – 273.15

Why do we have 273.15 in this relation, and not 273.16?

(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Answer:

The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K.

The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

The temperature 273.16 K is the triple point of water. It is not the melting point of ice.

The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.

Hence, absolute temperature (Kelvin scale) T is related to temperature tc, on Celsius scale as:

tc = (T – 273.15)

Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Both the temperatures can be related as:

(TF – 32)/ (180) = (TK – 273.15)/ (100)    … (i)

Let TF1 be the temperature on Fahrenheit scale and TK1 be the temperature on absolute scale. Both the temperatures can be related as:

(TF1 – 32)/ (180) = (TK1 -273.15)/ (100)    … (ii)

It is given that:

(TK1 – Tk) = 1K

Subtracting equation (i) from equation (ii) we get,

(TF1 – TF)/ (180) = (TK1 – Tk)/ (100) = (1/100)

(TF1 – TF) = (1 x 180)/ (100) = (9/5)

Triple point of water = 273.16 K

∴Triple point of water on absolute scale = (273.16) x (9/5) = 491.69

Question 5.

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Temperature             Pressure thermometer A                Pressure thermometer B

Triple-point of water    1.250 × 105 Pa                          0.200 × 105Pa

Normal melting point    1.797 × 105 Pa                          0.287 × 105 Pa

of sulphur

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?

(b) What do you think is the reason behind the slight difference in answers of thermometers A and B?

(The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Answer:

Given:

Triple point of water, T = 273.16 K.

At this temperature, pressure in thermometer A, PA = 1.250 × 105 Pa

Let T1 be the normal melting point of sulphur.

At this temperature, pressure in thermometer A, P1 = 1.797 x 105 Pa

According to Charles’ law, we have the relation:

(PA/T) = (P1/T1)

Therefore, T1 = (P1T)/ (PA)

= (1.797 x 105) x 273.16)/ (1.250 × 105)

= 392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as thermometer A is 392.69 K.

At triple point 273.16 K, the pressure in thermometer B, PB = (0.200 × 105) Pa

At temperature T1, the pressure in thermometer B, P2 = (0.287 × 105) Pa

According to Charles’ law, we can write the relation:

(PB/T) = (P1/T1)

= (0.200 x 105)/ (273.16) = (0.287 x 105)/ (T1)

Therefore T1 = ((0.287 x 105)/ (0.200 x 105)) x (273.16)

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases.

Hence, there is a slight difference between the readings thermometers A and B.

To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions.

At low pressure, these gases behave as perfect ideal gases.

Question 6.

A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C.

The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C.

What is the actual length of the steel rod on that day?

What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10–5 K–1

Answer:

Actual length of the steel rod l0 = 63 cm

Coefficient of linear expansion of steel, α = 1.20 × 10–5 K-1

The change in length is given as =Δl

ΔT= (T2-T1) = (45-27) °C = 18°C

Δl = (α l0 ΔT)

Therefore

Δl = (1.2 x10-5 x63 x18) cm

= 0.0136 cm

Hence, the actual length of the steel rod measured by the steel tape at 45  can be calculated as:

(l0 + Δl) = (63 + 0.0136) cm = 63.0136 cm (up to 3 significant places)

Therefore, the actual length of the rod at 27°C is 63.0 cm.

Question7.

A large steel wheel is to be fitted on to a shaft of the same material.

At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm.

The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft?

Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.

Answer:

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)

= 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.

Question 8.

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C.

What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.

Answer:

Initial temperature, T1 = 27.0°C

Diameter of the hole at T1, d1 = 4.24 cm

Final temperature, T2 = 227°C

Diameter of the hole at T2 = d2

Co-efficient of linear expansion of copper, αCu= 1.70 × 10–5 K–1

For co-efficient of superficial expansion β, and change in temperature ΔT, we have the relation:

(Change in area (ΔA))/ (Original Area (A) = β ΔT

(π (d22/4) – π (d12/4))/ (π (d12/4)) = (ΔA)/ (A)

(ΔA)/ (A) = (d22 - d21)/ (d21)

β = 2α

(d22 - d21)/ (d21) = 2 α ΔT

((d22 / d21) – 1) = 2 α (T2 – T1)

((d22) / (4.24)2) = 2 x 1.7 x 10-5(227-21) +1

(d22) = 17.98 + 1.0068 =18.1

Therefore,

d2 = 4.2544cm

Change in diameter = (d2 – d1) = (4.2544 – 4.24) = 0.0144 cm

Hence, the diameter increases by 1.44 × 10–2 cm.

Question 9.

A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports.

If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm?

Co-efficient of linear expansion of brass = 2.0 × 105 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.

Answer:

Initial Temperature = 27°C

Length of the brass wire at T1, l = 1.8 m

Final temperature, T2 = –39°C

Diameter of the wire, d = 2.0 mm = 2 × 10–3 m

Tension developed in the wire = F

Coefficient of linear expansion of brass, α= 2.0 × 10–5 K–1

Young’s modulus of brass, Y = 0.91 × 1011 Pa

Young’s modulus is given by the relation:

Y = (Stress)/ (Strain)

= (F/A)/ (ΔL/L)

ΔL = (F x L)/ (A x Y)

Where,

F = Tension developed in the wire

A = Area of cross-section of the wire.

ΔL = Change in the length, given by the relation:

ΔL = αL (T2 – T1) … (ii)

Equating equations (i) and (ii), we get:

αL (T2 – T1) = (FL)/( π(d/2)2 x Y )

F = α (T2 – T1) Y π (d/2)2

F= 2x10-5x (-39x-27) x3.14x0.91x1011x ((2 x 103)/ (2)) 2

=3.8 x 102 N

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is 3.8 ×102 N.

Question 10.

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter,

what is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C?

Is there a ‘thermal stress’ developed at the junction?

The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).

Answer:

Initial temperature, T1 = 40°C

Final temperature, T2 = 250°C

Change in temperature, ΔT = (T2 – T1) = 210°C

Length of the brass rod at T1, l1 = 50 cm

Diameter of the brass rod at T1, d1 = 3.0 mm

Length of the steel rod at T2, l2 = 50 cm

Diameter of the steel rod at T2, d2 = 3.0 mm

Coefficient of linear expansion of brass, α1 = 2.0 × 10–5 K–1

Coefficient of linear expansion of steel, α2 = 1.2 × 10–5 K–1

For the expansion in the brass rod, we have:

(Change in length (Δ l1)) / (Original length (l1)) = α1 ΔT

Therefore,

Δl1 = (50 x (2.1 x 10-5) x 210)

=0.2205cm

For the expansion in the steel rod, we have:

(Change in length (Δ l2)) / (Original length (l2)) = α2 ΔT

Therefore,

Δl2 = (50 x (1.2 x 10-5) x 210)

=0.126cm

Total change in the lengths of brass and steel,

Δl = Δl1 + Δl2

= 0.2205 + 0.126

= 0.346 cm

Total change in the length of the combined rod = 0.346 cm

No thermal stress is developed so the ends of rod are free to expand.

Question 11.

The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature?

Answer:

Given:

Coefficient of volume expansion of glycerine, αV = 49 × 10–5 K–1

Rise in temperature, ΔT = 30°C

Fractional change in its volume = (ΔV)/ (V)

This change is related with the change in temperature as:

(ΔV)/ (V) = (αV ΔT)

(VT2 – VT1) = (VT1αV ΔT)

(m/ ρT2) – (m/ ρT1) = (m/ ρT1) (αV ΔT)

Where, m = Mass of glycerine, ρT1 = Initial density at T1, ρT2 = Final density at T2

T1 - ρT2)/ ( ρT2) = (αV ΔT)

Where (ρT1 - ρT2) = Fractional change in density

Therefore, Fractional change in density of glycerine = 49 × 105 x 30

=1.47 x 10-2

Question 12.

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg.

How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings.

Specific heat of aluminium = 0.91 J g–1 K–1.

Answer:

Power of the drilling machine, P = 10 kW = (10 × 103) W

Mass of the aluminium block, m = 8.0 kg = (8 × 103) g

Time for which the machine is used, t = 2.5 min = (2.5 × 60) = 150 s

Specific heat of aluminium, c = 0.91 J g–1 K–1

Rise in the temperature of the block after drilling = δT

Total energy of the drilling machine = Pt

= (10 × 103 × 150)

= (1.5 × 106) J

It is given that only 50% of the power is useful.

Useful Energy ΔQ = (50/100) x 1.5 x 106

= (7.5 x 105) J

Also ΔQ = mcΔT

Therefore ΔT = (ΔQ)/ (mc)

= (7.5 x 105)/ ((8 x 103) x 0.91)

= 1030C

Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

Question 13.

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block.

What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1;

Heat of fusion of water = 335 J g–1).

Answer:

Mass of the copper block, m = 2.5 kg = 2500 g

Rise in the temperature of the copper block, Δθ = 500°C

Specific heat of copper, C = 0.39 J g–1 °C–1

Heat of fusion of water, L = 335 J g–1

The maximum heat the copper block can lose, Q = mCΔθ

= (92500 × 0.39 × 500)

= 487500 J

Let m1 g = amount of ice that melts when the copper block is placed on the ice block.

The heat gained by the melted ice, Q = m1L

Therefore, m1 = (Q/L)

= (487500)/ (335)

=1455.22g

Hence, the maximum amount of ice that can melt is 1.45 kg.

Question 14.

In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper

calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C.

Compute the specific heat of the metal.

If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Answer:

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T1 = 150°C

Final temperature of the metal, T2 = 40°C

Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g

Volume of water, V = 150 cm3

Mass (M) of water at temperature T = 27°C: 150 × 1 = 150 g

Fall in the temperature of the metal:

ΔT = (T1 – T2) = (150 – 40) = 110°C

Specific heat of water, Cw = 4.186 J/g/°K

Specific heat of the metal = C

Heat lost by the metal, θ = m CΔT … (i)

Rise in the temperature of the water and calorimeter system:

ΔT = (40 – 27) = 13°C

Heat gained by the water and calorimeter system:

Δθ’’ = m1CwΔT’

= (M + m′) Cw ΔT’ … (ii)

Heat lost by the metal = Heat gained by the water and calorimeter system

C m ΔT = (M + m’) Cw ΔT’

(200 × C × 110) = (1(50 + 25) × 4.186 × 13)

(175 x 4.186x13)/ (110x200) = 0.43Jg-1K-1

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Question 15.

Given below are observations on molar specific heats at room temperature of some common gases.

Gas                                    Molar specific heat (Cv) (cal mo1–1 K–1)

Hydrogen                                 4.87

Nitrogen                                  4.97

Oxygen                                    5.02

Nitric oxide                             4.99

Carbon monoxide                    5.01

Chlorine                                  6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases.

Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference.

What can you infer from the somewhat larger (than the rest) value for chlorine?

Answer:

The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion).

Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion.

Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.

If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = (5/2) R

= (5/2) x 1.98

= 4.95 cal mol-1k-1

With the exception of chlorine, all the observations in the given table agree with

(5/2)R

Question 16.

Answer the following questions based on the P-T phase diagram of carbon dioxide:

(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?

(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?

(c) What are the critical temperature and pressure for CO2? What is their significance?

(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?

Answer: • The solid, liquid and vapour phases of CO2 can exist in equilibrium at its triple point O i.e. temperature = 56.60C and atmospheric pressure = 5.11atm.
• From the vaporisation curve (1) and fusion curve (2) we can say that the both fusion and boiling points of CO2, decrease with the decrease of pressure.
• For CO2 the critical temperature is 31.10C and critical pressure is 73.0 atm. above its critical temperature (i.e. 31.10C), CO2 cannot be liquefied however large pressure may be applied.
• From (P-T) phase diagram we can say that:

CO2 is gaseous at –70°C, under 1 atm pressure

CO2 is solid at –60°C, under 10 atm pressure

CO2 is liquid at 15°C, under 56 atm pressure

Question 17.

Answer the following questions based on the P – T phase diagram of CO2:

(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?

(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?

(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm  pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.

(d) CO2 is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do you expect to observe?

Answer:

• No it does not go through a liquid phase. At 1 atm pressure and at –60°C, CO2 lies to the left of –56.6°C (triple point C).
• Hence, it lies in the region of vaporous and solid phases. Thus, CO2 condenses into the solid state directly, without going through the liquid state.
• It condenses directly to liquid phase. At 4 atm pressure, CO2 lies below 5.11 atm (triple point C).
• Hence, it lies in the region of vaporous and solid phases. Thus, it condenses into the solid state directly, without passing through the liquid state.
• The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves. When the temperature of a mass of solid CO2 (at 10 atm pressure and at –65°C) is increased, it changes to the liquid phase and then to the vaporous phase.
• It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.
• It departs from ideal gas behaviour as pressure increases.
• If CO2 is heated to 70°C and compressed isothermally, then it will not exhibit any transition to the liquid state. This is because 70°C is higher than the critical temperature of CO2. It will remain in the vapour state, but will depart from its ideal behaviour as pressure increases. Question 18.

A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body.

If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug?

Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg.

The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.

Answer:

Given:

Initial temperature of the body of the child, T1 = 101°F

Final temperature of the body of the child, T2 = 98°F

Change in temperature, ΔT = [(101 -98) x (5/9)] °C

Time taken to reduce the temperature, t = 20 min

Mass of the child, m = 30 kg = 30 × 103 g

Specific heat of the human body = Specific heat of water = c

= 1000 cal/kg/ °C

Latent heat of evaporation of water, L = 580 cal g–1

The heat lost by the child is given as:

∆θ = mc∆T

= (30 x 1000 x (101 -98)) x (5/9)

= 50000cal

Let m1 mass of the water evaporated from the child’s body in 20 min.

Loss of heat through water is given by:

∆θ = m1L

Therefore, m1 = (∆θ)/ (L)

= (50000)/ (580)

=86.2g

Therefore, Average rate of extra evaporation caused by the drug = (m1/t)

= (86.2)/ (200)

=4.3 g/min

Question 19.

A ‘thermoCol’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular.

A cubical icebox of side 30 cm has a thickness of 5.0 cm.

If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h.

The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermoCol is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]

Answer:

Side of the given cubical ice box, s = 30 cm = 0.3 m

Thickness of the ice box, l = 5.0 cm = 0.05 m

Mass of ice kept in the ice box, m = 4 kg

Time gap, t = 6 h = 6 × 60 × 60 s

Outside temperature, T = 45°C

Coefficient of thermal conductivity of thermoCol, K = 0.01 J s–1 m–1 K–1

Heat of fusion of water, L = 335 × 103 J kg–1

Let m’ = total amount of ice that melts in 6 h.

The amount of heat lost by the food:

θ = (KA (T1 – T2)/t)/d

Where,

A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3

θ = (0.01 x 0.54 x (45) x 6 x 60 x 60)/ (0.05)

=104976 J

But θ = m’L

Or m’ = (θ /L)

= (104976)/ (335 x 103)

=0.313Kg

Mass of ice left = (4 – 0.313) = 3.687 kg

Hence, the amount of ice remaining after 6 h is 3.687 kg.

Question 20.

A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove.

Estimate the temperature of the part of the flame in contact with the boiler.

Thermal conductivity of brass = 109 J s–1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.

Answer:

Given:

Thermal conductivity of brass K = 109 J s–1 m–1 K–1

Area of brass boiler, A = 0.15m2

Thickness d = 1.0cm = 10-2m

Boiling rate of water, R =0.6 kg/min

Time, t = 1 min = 60sec

Heat of vaporisation of water = 2256 × 103 J kg–1

The amount of heat flowing into water through the brass bas of the boiler is given as:

θ = (KA (T1 – T2)/t)/d   … (i)

Temperature of the flame in contact with the boiler= T1

Boiling point of water T2= 100°C

Therefore, heat required for boiling water, θ = mL ... (ii)

Equating equations (i) and (ii)

=> mL =(KA (T1 – T2)/t)/d

(T1 – T2) = (mLd)/ (KAt)

Therefore, (T1 – 100) = (6 x 2556 x 103 x 0.01)/ (109 x 0.15 x 60)

(T1 – 100) = 137.98

= 1380C

0r = 138 + 100

= 238 0C

Therefore, the temperature of the part of the flame in contact with the boiler is

238 0C.

Question 21.

Explain why:

(a) a body with large reflectivity is a poor emitter

(b) a brass tumbler feels much colder than a wooden tray on a chilly day

(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value

for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace

(d) the earth without its atmosphere would be inhospitably cold

(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Answer:

• 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.
• Brass is a good conductor of heat. When one touches a brass tumbler, heat is transferred from the body to the brass tumbler easily.
• Therefore, we feel cold by touching the brass tumbler on a chilly day. 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 cold.
• The temperature of red hot iron in the furnace is given as E= σ T4.
• When the iron is taken out in the open temperature T0 then its radiation energy is given as E= σ (T4 – T0) where E = Energy radiation

T = Temperature of optical, pyrometer To = Temperature of open

σ = Constant .Thus pyrometer measures the low value of the temperature of the red hot iron piece in the open.

• The lower layers of earth’s atmosphere reflect infrared radiations from earth back to the surface of earth.
• Thus the heat radiation received by the earth from the sun during the day is kept trapped by the atmosphere. If atmosphere of earth were not there, its surface would become cold to live.
• Because steam contains more heat in the form of latent (540 cal/g) than water.

Question 22.

A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Answer:

According to Newton’s law of cooling, we have:

Where

(-dT/dt) = K (T – To)

(dt)/ (K (T – To) = -K dt

Where

Temperature of the body = T

Temperature of the surroundings = T0 = 20°C

K is a constant

Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

Integrating equation (i), we get:

3080 (dt)/ (K (T – To)) =0300 K dt

[Loge (T – To (50)) 80 = -K[t] 0300

(2.3026)/ (K) Log10 ((80-20)/ (50-20)) =-300

(-2.3026)/ (300) log10 2 =K     ... (ii)

The temperature of the body falls from 60°C to 30°C in time = t’

Hence we get:

(2.3026)/ (K) Log10 ((60-20)/ (30-20)) =-t’

(-2.3026)/ (t’) log 10 4 = K … (iii)

Equating equations (ii) and (iii), we get:

(-2.3026)/ (t’) log 10 4 = (-2.3026)/ (300) log10 2

t' = (300 x 2) = 600s = 10min

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

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