Thermal Properties of Matter

Answer the following questions based on the P-T phase diagram of carbon dioxide: (c) What are the critical temperature and pressure for Carbon dioxide? What is its significance? (d) Is Carbon dioxide solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm?

Answer: c) As a result, the critical temperature and pressure of carbon dioxide are 31.1 degrees Celsius and 73.0 atmospheres, respectively. Carbon dioxide will not liquefy above this temperature,...

read more

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 CO_2 co-exist in equilibrium? (b) What is the effect of the decrease of pressure on the fusion and boiling point of CO_2?

Answer: (a) The liquid and vapor phases of $CO_2$ coexist at the triple point temperature and pressure of − 56.6 degrees Celsius and 5.11 atmospheres.   (b) When the pressure of $CO_2$...

read more

Explain why : (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

Answer: (c) The radiation energy emitted by a red hot iron piece placed in a furnace can be calculated using the equation E=σT4...

read more

A brass boiler has a base area of 0.15 m^2 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.

Solution: Given: A= 0.15 $m^2$ is the surface area of the brass boiler's base. D is the boiler's thickness, which is equal to 1 cm=0.01 m. The rate at which the brass boiler boils water is R=6.0...

read more

A ‘thermacole’ icebox is a cheap and an 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 the 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]

Solution: Given: Side of the cubical icebox, s =30 cm=3 x 10-2 m Thickness of the icebox, L =5.0 cm=0.05 m Mass of ice kept in the icebox, m=4 kg Time, t=6 h=6×60×60 = 21600 Outside temperature, T1=...

read more

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 minutes, 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 the latent heat of evaporation of water at that temperature is about 580 cal g–1.

Solution: Given: The initial temperature of the child, Ti =101o F The final temperature of the child, Tf=98oF Decrease in the temperature, △T=(101−98)= 30 F = 3×(5/9)= 1.670 C Mass of the child,...

read more

Given below are observations on molar specific heats at room temperature of some common gases.Generally, the specific heat of a monoatomic gas is 2.92 cal (mol K )‑1 , which is significantly lower from the specific heat of the above gases. Explain. It can be observed that chlorine has little larger value of specific heat, what could be the reason?

Answer: The diatomic molecules in the preceding list can move in translation, vibration, and rotation. A monoatomic gas can only move translationally. Heat is required to increase translational,...

read more

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 the specific heat of the metal?

Solution: Given the information in the question: Mass of the metal block, m = 0.20 kg = 200 g The initial temperature of the metal block, T1​ = 1500C The final temperature of the metal block, T2​ =...

read more

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.

Solution: Given: Power = 10 kW Mass of the small aluminium block, m = 8 kg = 8 x 103 g Time = 2.5×60=150 s Specific heat of aluminium, c= 0.91 J g–1 K–1. Total energy =P×t=104 ×150=15×105 J As 50%...

read more

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.

Solution: Given: Length of the brass rod = length of the steel rod = L0 = 50 cm Diameter of the brass rod = diameter of the steel rod = 3 mm Initial temperature ( T1) = 40°C Final temperature ( T2)...

read more

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 × 10–5 K–1; Young’s modulus of brass = 0.91 × 1011 Pa.

Answer: Given: Initial temperature, T1=27oC Length of the brass wire at 27oC, l =1.8m Final temperature, T2=- 390C Diameter of the wire, d=2.0mm=2×10−3m Coefficient of linear expansion of...

read more

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

Solution: Given the information in the question: Temperature, T= 27 °C The outer diameter of the shaft at 27 °C is d1 = 8.70 cm The diameter of the central hole in the wheel at 27 °C is d2 = 8.69 cm...

read more

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

Solution: Given: The temperature at which tape is calibrated: T = 27 °C At the temperature of 27 °C, the length of the tape is 1m = 100 cm T1 = 45 °C The length of the tape is 63 cm when the...

read more

Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: a) What is the absolute temperature of the 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?

Solution: (a) Given: Triple point of water, T= 273.16K. Pressure in thermometer A at the  triple point, PA =1.25×105 Pa Normal melting point of sulphur = T1 Pressure in thermometer A at this...

read more

(c) The absolute temperature (Kelvin scale) \mathbf{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?

Solution: (iii) The relationship exists because the melting point of ice equates to $273.15 K$ on the Kelvin scale, and the triple point of water corresponds to $273.16 K$ on the Kelvin scale. (iv)...

read more

(a) The triple-point of water is a standard fixed point in modern thermometer. 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 C elsius scale as mentioned above which were assigned the number 0^{\circ} \mathrm{C} and 100^{\circ} \mathrm{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 \mathrm{~K} . What is the other fixed point on this (Kelvin) scale?

Solution: i) Due to the fact that the melting and boiling points of water fluctuate with changes in pressure, the temperature of the triple point of water is not considered to be a standard fixed...

read more

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: \mathbf{R}=\mathbf{R}_{\mathrm{o}}\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_{0}\right)\right] The resistance is 101.6 \Omega at the triple-point of water 273.16 \mathrm{~K}, and 165.5 \Omega at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 \Omega ?

Solution: Given information in the question, Triple point temperature, $T_{0}=273.16 \mathrm{~K}$ Resistance at the triple point, $R_{0}=101.6 \Omega$ Normal melting point of lead, $T_{1}=600.5...

read more