Chapter 11 Thermal Properties of matter Solutions
Question - 11 : - The coefficient of volume expansion of glycerine is 49 x 10-5K-1. What is the fractional change in its density for a 30 °C rise in temperature ?
Answer - 11 : -
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 - 12 : - Power = 10 kW = 104 W
Mass, m=8.0 kg = 8 x 103 g
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 is 0.39 Jg-1°C-1. Heat of fusion of water = 335 Jg-1.
Answer - 13 : -
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 - 14 : - Mass of metal block, m = 0.20 kg = 200 g
Fall in the temperature of metal block,
ΔT = (150 – 40) °C = 110 °C
If C be the specific heat of metal, then heat lost by the metal block = 200 x Cx 110 cal Volume of water = 150 cm3
mass of water = 150 g
Increase in temperature of water = (40 – 27) °C = 13°C
Heat gained by water = 150 x 13 cal Water equivalent of calorimeter, w = 0.025kg = 25g
Heat gained by calorimeter,
Question - 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 mono atomic gases. Typically, molar specific heat of a mono atomic 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 - 15 : - The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat =5/2 R = 5/2 x 1.98 = 4.95, which agrees fairly well with all observations listed in the table except for chlorine. A mono atomic gas molecule has only the translational motion. A diatomic gas molecule, apart from translational motion, the vibrational as well as rotational motion is also possible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, heat is to be supplied to increase not only translational energy but also rotational and vibrational energies. Hence, molar specific heat of a diatomic gas is greater than that for mono atomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. shows that for chlorine molecule, at room temperature vibrational motion also occurs along with translational and rotational motions, whereas other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This is the reason that chlorine has somewhat larger value of molar specific heat.
Question - 16 : - (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 1
(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 - 16 : - (a) At the triple point, temperature = – 56.6 °C and pressure = 5.11atm.
(b) Both the boiling point and freezing point of CO2 decreaseif pressure decreases.
(c) The critical temperature and pressure of CO2 are 31.1°C and73.0 atm respectively. Above this temperature, CO2 will notliquefy/even if compressed to high pressures.
(d) (i) The point (- 70 °C, 1.0 atm) lies in the vapour region. Hence, CO2 isvapour at this point.
(ii) The point (- 60 °C, 10 atm) lies in the solid region. Hence, CO2 issolid at this point.
(iii) The point (15 °C, 56 atm) lies in the liquid region. Hence, CO2 isliquid at this point.
Question - 17 : - Answer the following questions based on the P – T phase diagram of CO2 (Fig. of question 17 given above)
(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 at 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 - 17 : -
(a) No, the CO2 doesnot go through the liquid phase. The point (1.00 atm, – 60°C) is to the lift ofthe triple-point O and below the sublimation curve OA. Therefore, when CO2 iscompressed at this point at constant temperature, the point moves perpendicularto the temperature-axis and enters the solid phase region. Hence, the CO2 vapourcondenses to solid directly without going through the liquid phase.
(b) CO2 at 4.0 atm pressure and room temperature (say, 27 °C)is in vapour phase. This point (4.0 atm, 27°C) lies below the vaporation curveOC and to the right of the triple point O. Therefore, when CO2 iscooled at this point at constant pressure, the point moves perpendicular to thepressure-axis and enters the solid phase region. Hence, the CO2 vapourcondenses directly to solid phase without going through the liquid phase.
(c) When the solid CO2 at – 65 °C is heated at 10 atm pressure,it is first converted into liquid. A further increase in its temperature bringsit into the vapour phase. If a horizontal line at P = 10 atm is drawn parallelto the T-axis, then the points of intersection of line with the fusion andvaporization curve give the fusion and boiling points at 10 atm.
(d) Above 31.1°C, the gas cannot be liquefied. Therefore, on being compressedisothermally at 70°C, there will be no transition to the liquid region.However, the gas will depart, more and more from its perfect gas behaviour withthe increase in pressure.
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 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 latent heat of evaporation of water at that temperature is about 580 cal g-1.
Answer - 18 : -
Question - 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 coefficient of thermal conductivity of thermacole is 0.01 Js-1 m-1 °C-1 [Heat of fusion of water = 335 x 103 J kg-1 ].
Answer - 19 : - Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m.Therefore, the total surface area’ of the icebox exposed to outside air is A =6 x (0.30 m)2 = 0.54 m2. The thickness of the iceboxis d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperaturedifference T1 – T2 = 45°C – 0°C = 45°C.
.•. Total heat entering the icebox in 6 h is given by
Question - 20 : - A brass boiler has a base area 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 Js-1 m-1 K-1.(Heat of vaporization of water = 2256 x 103 J kg-1 )
Answer - 20 : -
