Thermal Properties Of Matter

Given,

Triple point of Neon (T_Neon) = 23.57 K

Triple point of Carbon dioxide (T_{Carbon dioxide}) = 216.55 K

We know,

T_c = T_k – 273.15

Where, T_c is temperature in Celsius scale

T_k is temperature in Kelvin scale

Also,

T_f = (9/5)T_c + 32

Where, T_f is temperature in Fahrenheit scale

For Neon:

T_k = 24.57 K

∴ T_c = T_k – 273.15

⇒ T_c = 23.57 – 273.15

⇒ T_c = -248.58 °C

And T_f = (9/5)T_c + 32

⇒ T_f = (9/5)(-248.58) + 32

⇒ T_f = -447.44 + 32

⇒ T_f = -415.44°F

For Carbon dioxide:

T_k = 216.55 K

∴ T_c = T_k – 273.15

⇒ T_c = 216.55 – 273.15

⇒ T_c = -56.6 °C

And T_f = (9/5)T_c + 32

⇒ T_f = (9/5)(-56.6) + 32

⇒ T_f = -101.88 + 32

⇒ T_f = -69.88 °F

NOTE: The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.

Given,

Triple point of water in absolute scale A, T_A = 200A

Triple point of water in absolute scale B, T_B = 350B

We know that the triple point of water in Kelvin scale (T_K) is 273.15K.

So, 200A = 273.15

⇒ A = 273.15/200 …………(1)

Also, 350B = 273.15

⇒ B = 273.15/350 …………(2)

Dividing equation (1) by (2),we get

A/B = (273.15/200)/(273.15/350)

⇒ A/B = 350/200

⇒ A/B = 7/4

⇒ 4A = 7B

Hence, A and B are related by the relation 4A = 7B.

NOTE: The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.

Given,

Triple point of water, T₀ = 273.16 K

Resistance at triple point, R₀ = 101.6 Ω

Normal melting point of lead, T = 600.5 K

Resistance at normal melting point of lead, R = 165.5 Ω

Also,

R = R₀[1+α(T-T₀)]

Where, R₀ is the initial resistance

Ris the final resistance

T₀ is the initial temperature

T is the final temperature

∴ 165.5 Ω = (101.6 Ω)[1 + α(600.5 K - 273.16 K)]

⇒ 1+α(327.34) = 165.5/101.6

⇒ 1+327.34α = 1.629

⇒ 327.34α = 0.629

⇒ α = 0.629/327.34

⇒ α = 1.92 × 10^-3 K^-1

For R = 123.4 Ω,

123.4 Ω = (101.6 Ω)[1 + (1.92 × 10^-3 K^-1)(T - 273.16 K)]

⇒ 1 + (1.92 × 10^-3 K^-1)(T - 273.16 K) = 123.4/101.6

⇒ 1 + (1.92 × 10^-3 K^-1)(T - 273.16 K) = 1.214

⇒ (1.92 × 10^-3 K^-1)(T - 273.16 K) = 0.214

⇒ T - 273.16 K = 0.214/(1.92 × 10^-3 K^-1)

⇒ T – 273.16 K = 111.75 K

⇒ T = 111.75 K + 273.16 K

⇒ T = 384.91 K

(a) The melting point of ice and boiling point of water are temperature and pressure dependent but the triple point of water is not. Hence, the triple-point of water is a standard fixed point in modern thermometry.

(b) The absolute zero temperature or 0K is the other fixed point on the Kelvin scale.

(c) The temperature 273.16 K is the triple point of water. The temperature 273.15 K which is the melting point of ice in Kelvin scale corresponds to 0°C in Celsius scale. Hence, the relation has 273.15 in it and not 273.16.

(d) Let T_F be the temperature on Fahrenheit scale and T_K be the temperature on absolute scale.

The relation between both the temperatures is as follows:

(T_F - 32)/180 = (T_K - 273.15)/100 …………(1)

Let T_F’ be the temperature on Fahrenheit scale and T_K’ be the temperature on absolute scale.

The relation between both the temperatures is as follows:

(T_F’ - 32)/180 = (T_K’ - 273.15)/100 …………(2)

It is given that T_K’ – T_K = 1 K

Subtracting equation (1) from equation (2), we get

(T_F’ – T_F)/180 = (T_K’ - T_K)/100 = 1/100

⇒ T_F’ – T_F = (1×180)/100

⇒ T_F’ = 9/5

Triple point of water = 273.16 K

∴ Triple point of water on absolute scale, T = 273.16×(9/5)

⇒ T = 491.69

NOTE: The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.

(a) We know that the triple point of water, T₀ = 273.16 K.

At this temperature, pressure in thermometer A,

P_A = 1.250 × 10⁵ Pa

Let T be the normal melting point of sulphur.

At this temperature, pressure in thermometer A,

P = 1.797 × 10⁵ Pa

According to Charles’ law,

P_A/T₀ = P/T

∴ T = (PT₀)/P_A

⇒ T = (1.797×10⁵ Pa × 273.16 K)/(1.250×10⁵ Pa)

⇒ T = 392.69 K

Therefore, the absolute temperature corresponding to the normal melting point of sulphur according to the reading of thermometer A is 392.69 K.

At triple point T₀ = 273.16 K, the pressure in thermometer B,

P_B = 0.200×10⁵ Pa

At temperature T, the pressure in thermometer B,

P’ = 0.287 × 10⁵ Pa

According to Charles’ law,

P_B/T₀ = P’/T

⇒ (0.200 × 10⁵ Pa)/(273.16 K)= (0.287 × 10⁵ Pa)/T

∴ T = [(0.287 × 10⁵ Pa)/(0.200 × 10⁵ Pa)] × 273.16 K

⇒ T = 391.98 K

Therefore, the absolute temperature corresponding to the normal melting point of sulphur according to the reading of thermometer B is 391.98 K.

(b) The oxygen and hydrogen gases used in thermometers A and B respectively are not ideal gases. These gases have different physical behaviours. Hence, there is a slight difference between the readings of 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 ideal gases and will show same characteristics.

NOTE: Ideal gases are those gases which follow ideal gas equation exactly.

Given,

Length of the steel tape at temperature T = 27°C = 27 + 273.15 K = 300.15 K, L = 1 m

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

At temperature T₁ = 45°C = 45 + 273.15 K = 318.15 K, the length of the steel rod,

L₁ = 63 cm = 0.63m

Let L₂ be the actual length of the steel rod and L₂’ be the length measured by the steel tape at 45°C.

We know,

L₂’ = L + αL(T₁ - T)

∴ L₂’ = 1 m + (1.20 × 10^-5 K^-1)(1 m)(318.15 K – 300.15 K)

⇒ L₂’ = 1 + 0.000216

⇒ L₂’ = 1.000216 m = 100.0216 cm

Hence, the actual length of the steel rod at 45°C is given by

L₂ = (L₂’/L)×L₁

⇒ L₂ = [(100.0216 cm)/(100 cm)](63 cm) = 63.0136 cm Therefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.

Given,

Initial temperature, T₁ = 27°C = 27 + 273.15 = 300.15 K

Outer diameter of the steel shaft at T₁, D₁ = 8.70 cm

Diameter of the central hole in the wheel at T₁, d₁ = 8.69 cm

Coefficient of linear expansion of steel, α_steel = 1.20 × 10^-5 K^-1

Let the temperature be T₂ after the shaft is cooled using ‘dry ice’.

The wheel will slip on the shaft, if the change in diameter,

Δd = 8.69 – 8.70 = – 0.01 cm

Temperature T₂ can be calculated as follows.

Δd = D₁α_steel(T₂ – T₁)

⇒ -0.01 cm = 8.70 cm × 1.20 × 10^-5 K^-1 × (T₂ – 300.15 K)

⇒ (T₂ – 300.15 K) = -95.78 K

⇒ T₂ = 204.37 K = 204.37 – 273.15 = -68.78°C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is –68.78°C.

Given,

Initial temperature, T₁ = 27.0°C = 27 + 273.15 = 300.15 K

Diameter of the hole at T₁, d₁ = 4.24 cm

Final temperature, T₂ = 227°C = 227 + 273.15 = 500.15 K

Co-efficient of linear expansion of copper, α_Cu = 1.70 × 10^-5 K^-1

Let the diameter of the hole at T₂ be d₂.

If the co-efficient of superficial expansion is β, and the change in temperature is ΔT, then

Change in area (∆A)/Original area (A) = β∆T

∆A/A = [(πd₂²/4) - (πd₁²/4)]/(πd₁²/4)]

⇒ ∆A/A = (d₂² - d₁²)/d₁²

But β = 2α

∴ (d₂² - d₁²)/d₁² = 2α∆T

⇒ (d₂²/d₁²) -1 = 2α(T₂ – T₁)

⇒ d₂²/(4.24 cm)² – 1 = 2 × (1.7 × 10^-5 K^-1) (500.15 K – 300.15 K)

⇒ d₂²/(17.98 cm²) = 68 × 10^-4 + 1

⇒ d₂² = 17.98 cm² × 1.0068

⇒ d₂² = 18.1022 cm²

⇒ d₂ = 4.2546 cm

Change in diameter = d₂– d₁ = 4.2546 cm – 4.24 cm =0.0146 cm Hence, the diameter increases by 0.0146 cm.

Given,

Initial temperature, T₁ = 27°C = 27 + 273.15 K = 300.15 K

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

Final temperature, T₂ = –39°C = -39 + 273.15 K = 234.15 K

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

Co-efficient of linear expansion of brass, α = 2.0 × 10^–5 K^–1

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

Let the tension developed in the wire be T.

Young’s modulus is given by

Y = Stress/Strain

⇒ Y = (F/A)/(∆L/L)

⇒ ∆L = FL/AY

So, for tension T,

ΔL = TL/AY

Where, T = Tension developed in the wire

A = Area of cross-section of the wire

ΔL = Change in the length

But, ΔL = αL(T₂ – T₁)

∴ αL(T₂ – T₁) = TL/[π(d/2)² × Y]

⇒ T = απ(T₂ – T₁)Y(d/2)²

⇒ T = (2.0 × 10^–5 K^–1)×3.14×(234.15 K - 300.15 K)×( 0.91 × 10¹¹ Pa)×( 2 × 10^-3 m/2)²

⇒ T = -377.37 N

The negative sign indicates that the direction of tension is inwards.

Given,

Initial temperature, T₁ = 40°C = 40 + 273.15 K = 313.15K

Final temperature, T₂ = 250°C = 250 + 273.15 = 523.15 K

Length of the brass rod at T₁, L₁ = 50 cm = 0.5 m

Diameter of the brass rod at T₁, d₁ = 3.0 mm = 3 × 10^-3 m

Length of the steel rod at T₂, L₂ = 50 cm = 0.5 m

Diameter of the steel rod at T₂, d₂ = 3.0 mm = 3 × 10^-3 m

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

Change in temperature, ΔT = T₂ – T₁ = 523.15 K-313.15K=210 K

For the expansion in the brass rod,

Change in length (∆L₁)/Original length (L₁) = α₁ΔT

⇒ ΔL₁ = α₁L₁ΔT

⇒ ΔL₁ = (2.0 × 10^–5 K^–1)×(0.5 m)×(210 K)

⇒ ΔL₁ = 0.0021 m = 0.21 cm

For the expansion in the steel rod,

Change in length (∆L₂)/Original length (L₂) = α₂ΔT

⇒ ∆L₂ = α₂L₂ΔT

⇒ ∆L₂ = (1.2 × 10^–5 K^–1)×(0.5 m)×(210 K)

⇒ ∆L₂ = 0.00126 m = 0.126 cm

Total change in length, ΔL = ∆L₁ + ∆L₂

⇒ ΔL = 0.21 cm + 0.126 cm

⇒ ∆L= 0.336 cm

No thermal stress develops at the junction since both the ends of the rod expand freely.

Given,

Coefficient of volume expansion of glycerin, α_V = 49 × 10^–5 K^–1

Rise in temperature, ΔT = 30°C = 30 K

Fractional change in its volume = ΔV/V

We know,

ΔV/V = α_VΔT

⇒ V_T2 - V_T1 = V_T1α_VΔT

⇒ (m/ρT₂)-(m/ρT₁) = (m/ρT₁)α_VΔT

⇒ [(1/ρT₂)-(1/ρT₁)]/(1/ρT₁) = α_VΔT

⇒ (ρ_T1 - ρ_T2)/ρ_T2 = α_VΔT

Where, m is the mass of glycerin

ρ_T1 is the initial density at T₁

ρ_T2 is the initial density at T₂

(ρ_T1 - ρ_T2)/ρ_T2 is the fractional change in density

So, fractional change in density of glycerine = 49 × 10^–5 K^–1 × 30 K

= 1.47 × 10^-2

Given,

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

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

Machine operation time, t = 2.5 min = 2.5×60=150 s Specific heat of aluminium, c = 0.91 J g^–1 K^–1

Let the rise in the temperature of the block after drilling be ΔT.

Total energy of the drilling machine, E = Pt

⇒ E = 10 × 10³ × 150

⇒ E = 1.5 × 10⁶ J

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

So, Useful energy, ∆Q = (50/100) × E

⇒ ΔQ = 7.5 × 10⁵ J

But ∆Q = mc∆T

∴ ∆T = ∆Q/mc = (7.5 × 10⁵ J)/(8×10³ g × 0.91 J g^–1 K^–1) = 103 K Hence, the rise in the temperature of the block is 103 K in 2.5 minutes of drilling.

NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Given,

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

Rise in the temperature of the copper block, ΔT = 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 that the copper block can lose, Q = mCΔT

⇒ Q = 2500 g × 0.39 J g^–1 °C^–1 × 500 °C

⇒ Q = 487500 J

Let m’ be the amount of ice that melts when the copper block is placed on the ice block.

So, the heat gained by the melted ice, Q = m’L

∴ m’ = Q/L

⇒ m’ = 487500/335

⇒ m’ = 1455.22 g = 1.455 kg

Hence, the maximum amount of ice that can melt is 1.455 kg.

NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Given,

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 of water at temperature T = 27°C, M = 150 × 1 = 150 g

Fall in the temperature of the metal, ΔT = T₁ – T₂

⇒ ΔT = 150 – 40

⇒ ΔT = 110°C

Specific heat of water, C_w = 4.186 J g^-1 °C^-1

Let the specific heat of the metal be C.

Heat lost by the metal, θ = mCΔT … (1)

Rise in the temperature of the water and calorimeter system, ΔT′’ = 40 – 27

⇒ ΔT’’ = 13°C

Heat gained by the water and calorimeter system, Δθ′′ = m₁C_wΔT’ ⇒ Δθ′′= (M+m′)C_wΔT’ … (2)

We know that,

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

So, mCΔT = (M+m’)C_wΔT’

⇒ 200 g × C × 110 °C = (150 g + 25 g) × 4.186 J g^-1 °C^-1 × 13 °C

∴ C = (175 × 4.186 × 13)/(110 × 200)

⇒ C = 0.43 J g^-1 K^-1

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

NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

The gases given in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom. Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the degrees of freedom. 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 is equal to (5/2)R (where, R = Ideal gas law constant = 1.98 cal mol^-1 K^-1)

So, molar specific heat = (5/2) × 1.98 = 4.95 cal mol^-1 K^-1

With the exception of chlorine, all the observations in the given table agree with this value of molar specific heat. This is because of the fact that at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

NOTE: Molecular degrees of freedom refer to the number of ways a molecule in the gas phase may move, rotate, or vibrate in space.

The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

The P-T phase diagram of CO₂ is as follows:

(a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the triple point, i.e., the point corresponding to the temperature value of - 56.7° C and pressure value of 5.1 atm.

(b) With the decrease in pressure, both the fusion and boiling point of carbon dioxide will decrease.

(c) For carbon dioxide, the critical temperature is 31°C and critical pressure is 72.9 atm. If the temperature of carbon dioxide is more than 31° C, it cannot be liquified irrespective of the applied pressure.

(d) Carbon dioxide will be a vapour at 70°C under 1 atm, a solid at -6°C under 10 atm and a liquid at 15°C under 56 atm.

The P-T phase diagram of CO₂ is as follows:

(a) Since the temperature -60° C lies to the left of 56.7° C on the curve i.e., it lies in the region vapour and solid phase, carbon dioxide will condense directly into the solid without becoming liquid.

(b) Since the pressure 4 atm is less than 5.1 atm the carbon dioxide will condense directly into solid without becoming liquid.

(c) When a solid CO₂ at 10 atm pressure and -65° C temprature is heated, it is first converted into liquid. A further increase in temperature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the Temperature-axis, then the points of intersection of this line with the fusion and vaporization curve will give the fusion and boiling points of CO₂ at 10 atm.

(d) Since 70° C is higher than the critical temperature of CO₂, the CO₂ gas cannot be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will deviate from its perfect gas behaviour with increasing pressure.

Given,

Mass of the child, m = 30kg

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 = 3°F

⇒ ΔT = [3 × (5/9)] °C

⇒ ΔT = 1.667 °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^-1 °C^-1

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

The heat lost by the child is given as

∆θ = mc∆T

⇒ ∆θ = 30 kg × 1000 cal kg^-1 °C^-1 × 1.667 °C = 50000 cal

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

Loss of heat through water is given by

∆θ = m₁L

∴ m₁ = ∆θ/L

⇒ m₁ = (50000 cal)/(580 cal g^–1) = 86.2 g

Hence, average rate of extra evaporation caused by the drug, R_ave = m₁/t

⇒ R_ave = (86.2 g)/(200 min)

⇒ R_ave = 4.3 g/min.

NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Given,

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 = 21600 s

Outside temperature, T = 45°C

Coefficient of thermal conductivity of thermacole,

K = 0.01 J s^–1 m^–1 K^–1

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

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

The amount of heat lost by the food, θ = KA(T - 0)t/l

Where, A = Surface area of the box = 6×s×2 = 6×0.3×2 = 0.54 m³

∴ θ = [(0.01 Js^–1m^–1K^–1)×(0.54m³)×(45°C)×(21600s)]/(0.05 m)

⇒ θ = 104976 J

But θ = m'L

∴ m’ = θ/L = 104976/(335 × 10³ J kg^-1)

⇒ m’ = 0.313 g

Mass of ice left = 4 – 0.313 = 3.687 kg

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

NOTE: Heat of fusion is the amount of heat supplied to a specific quantity of the substance to change its state from a solid to a liquid at constant pressure.

Given,

Base area of the boiler, A = 0.15 m²

Thickness of the boiler, l = 1.0 cm = 0.01 m

Boiling rate of water, R = 6.0 kg/min

Mass, m = 6 kg

Time, t = 1 min = 60 s

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

Heat of vaporisation, L = 2256 × 10³ J kg^–1

The amount of heat flowing into water through the brass base of the boiler is given by

θ = KA(T₁ - T₂)t/l ....(1)

where, T₁ is the temperature of the flame in contact with the boiler

T₂ is the boiling point of water (= 100°C)

Heat required for boiling the water is given by

θ = mL … (2)

Equating equations (1) and (2), we get

mL = KA(T₁ - T₂)t/l

⇒ T₁ - T₂ = mLl/KAt

⇒ T₁ - T₂ = (6 kg × 2256 × 10³ J kg^-1× 0.01 m)/(109 J s^–1 m^–1 K^–1× 0.15 m²× 60 s)

⇒ T₁ - T₂ = 137.98°C

⇒ T₁ = 137.98°C + 100°C

⇒ T₁ = 237.98°C

So, the temperature of the part of the flame in contact with the boiler is 237.98°C.

NOTE: Heat of vaporisation is the amount of heat supplied to a specific quantity of the substance to change its state from a liquid to a gas at constant pressure.

A body with a large reflectivity is a poor absorber of light radiations since it doesn’t absorb energy. 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 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.

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 - T₀⁴)

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 = σ T⁴.

In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth’s surface. So, without its atmosphere, earth would be inhospitably cold.

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).

According to Newton’s law of cooling,

(-dT/T) = K(T - T₀)

⇒ dT/(T - T0) = -Kdt ………….(1)

Where, T is the temperature of the body

T₀ is the temperature of the surroundings = 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 (1), we get

⇒

⇒

⇒

⇒ ………….(1)

Let t’ be the time it takes to cool from 60 °C to 30 °C. Then,

⇒

⇒

⇒

⇒

⇒ t’ = 2 × 300

⇒ t’ = 600 s

⇒ t’ = 600/60 = 10 min

Hence, it takes 10 min to cool the body from 60 °C to 30 °C.

NOTE: Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).