11.12. 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 aluminum 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, ∆Q = (50/100) × 1.5 × 10⁶  =  7.5 × 10⁵ J
But ∆Q = mc∆T
∴ ∆T = ∆Q / mc
= (7.5 × 10⁵) / (8 × 10³ × 0.91)
= 103^o C
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

11.13. 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
= 2500 × 0.39 × 500
= 487500 J
Let m₁ g be the amount of ice that melts when the copper block is placed on the ice block.
The heat gained by the melted ice, Q = m₁L
∴ m₁ = Q / L  =  487500 / 335  =  1455.22 g
Hence, the maximum amount of ice that can melt is 1.45 kg.

11.14. 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, C_w = 4.186 J/g/°K
Specific heat of the metal = C
Heat lost by the metal, θ = mCΔT … (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′′ = m₁ C_wΔT^’
= (M + m′) C_w ΔT^’ … (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m^’) C_w ΔT^’
200 × C × 110 = (150 + 25) × 4.186 × 13
∴ C = (175 × 4.186 × 13) / (110 × 200)  =  0.43 Jg^-1K^-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

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

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?

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) × 1.98 = 4.95 cal mol^-1 K^-1
With the exception of chlorine, all the observations in the given table agree with (5/2)R.
This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

11.16. 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 P–T phase diagram for CO₂ is shown in the following figure:

11.17. 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

The P–T phase diagram for CO₂ is shown in the following figure:

(a) Since the temprature -60° C lies to the left of 56.6° C on the curve i.e. lies in the region vapour and solid phase, so carbon dioxide will condense directly into the solid without becoming liquid.

(b) Since the pressure 4 atm is less than 5.11 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 temprature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the T-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 temprature of CO_2, so the CO₂ gas can not be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will depart more and more from its perfect gas behaviour with the increase in pressure.

11.18. 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

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) × (5/9) ] ^o 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 × 1000 × (101 – 98) × (5/9)
= 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
= (50000 / 580)  =  86.2 g
∴ Average rate of extra evaporation caused by the drug = m₁ / t
= 86.2 / 200
= 4.3 g/min.

11.19. A ‘thermacole’ 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 thermacole 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 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 = 6s² = 6 × (0.3)² = 0.54 m³
θ = 0.01 × 0.54 × 45 × 6 × 60 × 60 / 0.05  =  104976 J
But θ = m’L
∴ m‘ = θ/L
= 104976/(335 × 10³)  =  0.313 kg
Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.

11.20. 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

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    ….(i)

where,
T₁ = Temperature of the flame in contact with the boiler
T₂ = Boiling point of water = 100°C
Heat required for boiling the water:
θ = mL … (ii)
Equating equations (i) and (ii), we get:
∴ mL = KA(T₁ – T₂) t / l
T₁ – T₂ = mLl / KAt
= 6 × 2256 × 10³ × 0.01 / (109 × 0.15 × 60)
= 137.98 ^o C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

11.21. 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) 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.

(b) 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.

(c) 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
T_o = 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

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

(e) 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).

11.22. 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:
(-dT/T) = K(T – T₀)
dT / K(T – T₀)  =  –Kdt            ….(i)
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: