{"id":39685,"date":"2022-12-07T13:05:13","date_gmt":"2022-12-07T07:35:13","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=39685"},"modified":"2022-12-07T13:08:04","modified_gmt":"2022-12-07T07:38:04","slug":"ts-inter-1st-year-physics-study-material-chapter-12","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-12\/","title":{"rendered":"TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material<\/a> 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.<\/p>\n

TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter<\/h2>\n

Very Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nDistinguish between heat and temperature. [TS Mar. ’15]
\nAnswer:
\nDifferences between heat and temperature :<\/p>\n\n\n\n\n\n\n\n\n
Heat<\/td>\nTemperature<\/td>\n<\/tr>\n
1. Heat is a form of energy.<\/td>\n1. It represents relative degree of hotness (or) Coldness of a body.<\/td>\n<\/tr>\n
2. Unit: joule (or) calorie<\/td>\n2. Unit: \u00b0C or \u00b0F<\/td>\n<\/tr>\n
3. It is cause<\/td>\n3. It is effect.<\/td>\n<\/tr>\n
4. Heat is measured by calorimeter<\/td>\n4. It is measured by thermometer.<\/td>\n<\/tr>\n
5. Quantity of heat supplied Q = m st<\/td>\n5. Change in temperature of a body \u2206 t = \\(\\frac{Q}{ms}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 2.
\nExplain triple point of water.
\nAnswer:
\nThe temperature of a substance remains constant during its change of state.<\/p>\n

A graph plotted between temperature (T) and pressure (P) of a substance during change of state is called “phase diagram”.
\n\"TS<\/p>\n

In phase diagram of water, the P – T plane is divided into three regions.<\/p>\n

The line ‘AO’ is called fusion curve. It gives equilibrium temperature and pressure between solid and liquid states.<\/p>\n

The line ‘CO’ is called vaporisation curve. It gives equilibrium temperature and pressure between liquid and vapour states.<\/p>\n

The line ‘BO’ is called sublimation curve. It gives the relation between of temperature and pressure between solid and vapour states.<\/p>\n

Triple Point:
\nAt point ‘O’ the curves AO, BO and CO will intersect.<\/p>\n

It gives the temperature and pressure at which solid, liquid and vapour states of water are in equilibrium.<\/p>\n

Coordinate of triple point of water a temperature = 273.16 K, pressure = 0.006 atmos (or) 611 pascals.<\/p>\n

Question 3.
\nWhat are the lower and upper fixing points in Celsius and Fahrenheit scales? [TS Mar. ’16; AP Mar. ’19, ’18 ’16, May ’14]
\nAnswer:
\nCentigrade (Celsius) scale of temperature:
\nIn centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0\u00b0C. The upper limit is boiling point of water at 1 atm pressure, as 100\u00b0C.<\/p>\n

Fahrenheit scale of temperature :
\nIn Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32\u00b0F. The upper fixed point is boiling point of water at 1 atm pressure, as 212\u00b0F.<\/p>\n

\"TS<\/p>\n

Question 4.
\nDo the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale?
\nAnswer:
\nYes. Coefficients of expansion \u03b1, \u03b1a<\/sub> and \u03b1v<\/sub> are not same in Celsius scale and in Fahrenheit scale. In Fahrenheit scale values of \u03b1, \u03b1a<\/sub> and \u03b1v<\/sub> are less than those in Celsius scale.<\/p>\n

Since magnitude of 1\u00b0C > magnitude of 1\u00b0F this change takes place. The values of Celsius scale are 1.8 times more than the values in Fahrenheit scale.
\n\u03b1F<\/sub> = 5\/9 \u03b1c<\/sub><\/p>\n

Question 5.
\nCan a substance contact on heating? Give an example. [AP Mar. ’18, ’16, May ’16; TS May, ’18, ’16]
\nAnswer:
\nYes. Some substances will contact on heating. Ex: Leather, rubber, cast Iron type metal.<\/p>\n

Question 6.
\nWhy gaps are left between rails on a railway track? [TS Mar. ’19; AP Mar. ’19, ’17, ’16, ’09; May ’16; June ’15]
\nAnswer:
\nTo allow the linear expansion rails.<\/p>\n

In summer temperature of atmosphere increases so rails will expand. If no gap is given between rails then the rails will bend it leads to accidents. If gap is given the rails will expand into that gap and that track is safe.<\/p>\n

Question 7.
\nWhy do liquids have no linear and areal expansions? [TS Mar. ’19]
\nAnswer:
\nLiquids have only volume expansion. No linear expansion or areal expansion. Because liquids does not have any independent shape they must be taken in a container. So we will consider only volume of liquid.<\/p>\n

Question 8.
\nWhat is latent heat of fusion? [AP & TS May ’17]
\nAnswer:
\nLatent heat of fusion (melting):
\nIt is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from solid to liquid or from liquid to solid.<\/p>\n

Question 9.
\nWhat is latent heat of vapourisation? [AP Mar. ’13]
\nAnswer:
\nLatent heat of vapourisation :
\nIt is defined as the amount of heat energy absorbed or rejected by unit mass of substance while converting from liquid to vapour or from vapour to liquid state.<\/p>\n

Question 10.
\nWhy are utensils coated black? Why is the bottom of the utensils made of copper? [AP May ’18; TS Mar. ’18]
\nAnswer:
\nLower portion of the utensils is in contact with fire. Black bodies are good heat absorbers. So, a black bottom will absorb more heat.<\/p>\n

Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.<\/p>\n

\"TS<\/p>\n

Question 11.
\nWhat is triple point of Water? Mention the values of temperature and pressure at triple point of water. [TS June ’15]
\nAnswer:
\nTriple point:
\nThe temperature and pressure where a substance can coexist in all its three states is called the “triple point”.<\/p>\n

i.e., The substance will exist as a solid, as liquid and as vapour at that particular temperature and pressure.<\/p>\n

For water the triple point is at a temperature of 273.16 K and at a pressure of 6.11 \u00d7 10-3<\/sup> atmospheres or nearly 610 pascals.<\/p>\n

Question 12.
\nState Boyle’s law and Charles law. [AP June 15; TS Mar. 15]
\nAnswer:
\nBoyle’s Law :
\nAt constant temperature, the volume (V) of a given mass of a gas is inversely proportional to its pressure (P).
\n\u2234 V \u221d \\(\\frac{1}{P}\\) \u21d2 PV = constant = K.<\/p>\n

Charles Law:
\nAt constant pressure, the volume (V) of a given mass of a gas is directly proportional to its absolute temperature (T).
\n\u2234 V \u221d T \u21d2 \\(\\frac{V}{T}\\) = K (constant)<\/p>\n

Question 13.
\nState Wein’s displacement law. [AP Mar. ’17]
\nAnswer:
\nWein’s Displacement Law :
\nThe wavelength (Ain) of maximum intensity of emission of black body radiation is inversely proportional to absolute temperature (T) of the black body.
\ni.e., \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\) (or) \u03bbm<\/sub> = \\(\\frac{b}{P}\\)
\nwhere ‘b’ is called ‘Wein’s constant”.<\/p>\n

Question 14.
\nVentilators are provided in rooms just below the roof. Why?
\nAnswer:
\nDensity of hot air is less. So in a room hot air goes to top layers i.e., nearer to the roof.<\/p>\n

When ventilators are provided nearer to the roof hot air will escape easily from room. So we feel that the room is cool and circulation of air will become easy.<\/p>\n

Question 15.
\nDoes a body radiate heat at 0 K? Does it radiate heat at 0\u00b0C?
\nAnswer:
\nAccording to Precost’s theory, every body above zero Kelvin will radiate heat energy to the surroundings.
\nSo, i) A body at O’ Kelvin does not radiate heat energy.
\nii) A body at 0\u00b0C i.e., at 273Kwill radiate heat energy.<\/p>\n

Question 16.
\nState the different modes of transmission of heat. Which of these modes require medium? [TS May ’18]
\nAnswer:
\nTransmission of heat is of three types.
\n1) Conduction 2) Convection 3) Radiation<\/p>\n

For propagation of heat energy medium is required in case of conduction and convection.<\/p>\n

\"TS<\/p>\n

Question 17.
\nDefine coefficient of thermal conductivity and temperature gradient.
\nAnswer:
\n“The coefficient of thermal conductivity”
\n(K) it is the quantity of heat flowing normally per second through unit area of the substance per unit temperature gradient.
\n\"TS<\/p>\n

‘Temperature gradient” is defined as the change in temperature along the conductor per unit length.<\/p>\n

Temperature gradient
\nChange in temperature
\n\"TS<\/p>\n

Question 18.
\nDefine emissive power and emissivity.
\nAnswer:
\nEmissive power:
\nThe emissive power of a body is defined as the energy radiated by the body per second per unit area at a given temperature and wavelength.<\/p>\n

Emissivity:
\nEmissivity is defined as the ratio of the emissive power of a body to that of a black body at the same temperature.<\/p>\n

Question 19.
\nIs there any substance available in nature that contracts on heating? If so, give an example. [TS May ’16]
\nAnswer:
\nYes. Some substances will contact on heating.
\nEx: Leather, rubber, cast Iron type metal.<\/p>\n

Question 20.
\nWhat is greenhouse effect? Explain global warming. [AP Mar. ’15, ’13; TS Mar. & May ’16]
\nAnswer:
\nGreen house effect: Earth will absorb heat radiation and reradiate heat energy of longer wavelength. This longer wave length heat radiation is reflected back to earth due to green house gases such as Carbon dioxide [CO2<\/sub>], Methane (CH4<\/sub>) Chloroflurocarbons, Ozone (O3<\/sub>), etc. As a result temperature of earth’s atmosphere is gradually increasing. This is known as “green house effect.”<\/p>\n

Global warming:
\nEarth receives heat energy during day time from sun. It reradiates heat energy in the form of longer electromagnetic waves.<\/p>\n

But due to presence of green house gases the longer electromagnetic waves were reflected back to earth. As a result temperature of earth’s atmosphere is gradually increasing.<\/p>\n

This process will increase with the increased content of green house gases in atmosphere. As a result temperature of earth\u2019s atmosphere increases gradually.<\/p>\n

Question 21.
\nDefine absorptive power of a body. What is the absorptive power of a perfect black body?
\nAnswer:
\nAbsorptive power of a body is defined as the ratio of energy absorbed by the body within the wave length range of A and A + dA to the total energy flux following on the body.
\nAbsorptive power,
\n\"TS<\/p>\n

Question 22.
\nState Newton’s law of cooling. [AP Mar. ’18, ’16, May ’18, ’17, June ’15; TS Mar. ’18, TS May ’16]
\nAnswer:
\nNewton’s Law of cooling states that the rate of loss of heat of a hot body is directly pro-portional to the difference in temperature between the body and its surroundings pro-vided the difference in temperatures is small and the nature of the radiating surface remains same.
\n\"TS
\nWhere k is the proportionality constant<\/p>\n

Question 23.
\nState the conditions under which Newton’s law of cooling is applicable. [AP May ’16; TS June ’15]
\nAnswer:
\nNewton’s law of cooling is applicable<\/p>\n

    \n
  1. loss of heat is negligible by conduction and only when it is due to convection.<\/li>\n
  2. loss of heat occurs in a streamlined flow of air i.e., forced convection.<\/li>\n
  3. temperature of the body is uniformly distributed over it.<\/li>\n
  4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.<\/li>\n<\/ol>\n

    \"TS<\/p>\n

    Question 24.
    \nThe roofs of buildings are often painted white during summer. Why? [TS Mar. ’17, ’15; AP May ’16]
    \nAnswer:
    \nWhen roofs of buildings are coated white we will feel relatively cold during summer.<\/p>\n

    Absorptive power of white surface is less. So roofs coated white will absorb less heat energy. So less quantity of heat is transmitted into the house. So we feel less hot or cold inside the house.<\/p>\n

    Question 25.
    \nWhat is thermal expansion? [TS Mar. ’16]
    \nAnswer:
    \nThe increase in interatomic distance due to thermal energy is called “thermal expansion”.<\/p>\n

    As a result the length solids or volume of liquids or pressure of gases will increase.<\/p>\n

    Question 26.
    \nWhy is it easier to perform the skating on the snow? [TS Mar. ’16]
    \nAnswer:
    \nDue to increase of pressure melting point decreases, So it is easier to perform the skating on the snow.<\/p>\n

    Short Answer Questions<\/span><\/p>\n

    Question 1.
    \nExplain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
    \nAnswer:
    \nCelsius (Centigrade) scale of temperature :
    \nIn centigrade scale of temperature lower fixed point is freezing point of water at one atmosphere pressure, as 0\u00b0C. The upper limit is boiling point of water at 1 atm pressure, as 100\u00b0C.<\/p>\n

    The interval between lower limit and upper limit [100 – 0 = 100] is divided into 100 equal parts and each part is called 1\u00b0C.<\/p>\n

    Fahrenheit scale of temperature :
    \nIn Fahrenheit scale, the lower fixed point is freezing point of water at one atmosphere pressure, as 32\u00b0F. The upper fixed point is boiling point of water at 1 atm pressure, as 212\u00b0F.<\/p>\n

    The interval between upper fixed point and lower fixed point (212 -32 = 180) is divided into 180 equal parts and each part is called 1\u00b0F.<\/p>\n

    Relation between Fahrenheit and Celsius scale of temperatures:
    \nIn both scales, lower limit and upper limit are same. The only change is in numerical values of lower and upper limits.<\/p>\n

    In Fahrenheit lower limit = 32, upper limit = 212, difference of limits = 180<\/p>\n

    In Celsius scale lower limit = 0, upper limit = 100, difference of limits = 100
    \n\"TS
    \nC = Temperature in Celsius scale.
    \nF = Temperature in Fahrenheit scale.<\/p>\n

    Question 2.
    \nTwo identical rectangular strips one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating?
    \nAnswer:
    \nWhen two dissimilar metals say copper and steel are riveted together that arrangement is called “bimetallic strip”.<\/p>\n

    When a bimetallic strip is heated copper strip will expand more than steel due to more expansion coefficient.
    \n\"TS<\/p>\n

    Since they are riveted they must expand as a common piece. As a result bimetallic strip will bend in the form of an arc. For the metallic strip with high a its length is more so it is on the outer side. For the strip with less a its length is less. It will be at the inner side of the arc.<\/p>\n

    \"TS<\/p>\n

    Question 3.
    \nPendulum clocks generally go fast in winter and slow in summer. Why? [TS Mar. ’19, ’17]
    \nAnswer:
    \nIn summer due to increase in temperature of atmosphere length of pendulum will increase.<\/p>\n

    Time period of pendulum T = 2\u03c0\\(\\sqrt{\\frac{l}{g}}\\)<\/p>\n

    T \u221d \u221al. So it will make less number of oscillations per day. So clock will run slowly in summer.<\/p>\n

    In winter temperature of atmosphere decreases. So length of pendulum decreases.<\/p>\n

    Hence time period of oscillation will also decrease. As a result, pendulum will make more oscillations per day so clocks will run fast in winter.<\/p>\n

    Question 4.
    \nIn what way is the anomalous behaviour of water advantageous to aquatic animals? [AP Mar. ’18, 17, 14; May 18. 17, 14; TS May ’18]
    \nAnswer:
    \nIn cold countries and at polar region temperature falls below 0\u00b0C at winter. So surface of water will be frozen. Due to anomalous expansion of water even though the surface of lakes, and sea are frozen water will exist at bottom layers at 4\u00b0C.<\/p>\n

    Different layers in between ice and bottom will have different temperatures like 1\u00b0C, 2\u00b0C or 3\u00b0C. In these layers, aquatic animals are able to survive even in winter.<\/p>\n

    Anomalous expansion of water helps for the survival of aquatic life at polar region and in cold countries.<\/p>\n

    Question 5.
    \nExplain conduction, convection and radiation with examples. [TS Mar. ’18, ’16, ’15, June ’15; AP Mar. ’19, ’15, May, ’16]
    \nAnswer:
    \nConduction :
    \nIt is a mode of transfer of heat from one part of the body to another, from particle to particle in the direction of fall of temperature without any actual movement of the heated particles.
    \nEx: When one end of a metal rod is heated, its other end becomes hot. Here, the heat goes from hot end of the metal rod towards cold end, by conduction.<\/p>\n

    Convection :
    \nIt is a mode of transfer of heat from one part of the medium to another part by the actual movement of the heated particles of the medium.
    \nEx : Seabreeze, Tradewind, etc.<\/p>\n

    Radiation :
    \nIt is a mode of transfer of heat from the source to the receiver without any actual movement of source or receiver and also without heating the intervening medium.
    \nEx : Heat from sun comes to us through radiation. On standing near fire, we feel hot as heat comes to us through radiation.<\/p>\n

    Long Answer Questions<\/span><\/p>\n

    Question 1.
    \nExplain thermal conductivity and coefficient of thermal conductivity.
    \nA copper bar of thermal conductivity 401 W (mK) has one end at 104\u00b0C and the other end at 24\u00b0C. The length of the bar is 0.10 m and the cross-sectional area is 1.0 \u00d7 10-6<\/sup> m-2<\/sup>. What is the rate of heat conduction along the bar?
    \nAnswer:
    \nThe ability to conduct heat in solids is called ‘Thermal conductivity.”<\/p>\n

    Consider a bar with rectangular cross-section as shown in the figure. The faces ABCD and EFGH are maintained at \u03b81<\/sub> and \u03b82<\/sub> respectively (\u03b81<\/sub> > \u03b82<\/sub>). Heat passes from one end to the other.
    \n\"TS<\/p>\n

    The amount of heat conducted (Q) depends on,<\/p>\n

      \n
    1. Amount of heat conducted Q is proportional to area of cross-section A perpendicular to flow.
      \n\u2234 Q \u221d A ………. (1)<\/li>\n
    2. is proportional to temperature difference between the two ends.
      \n\u2234 Q \u221d (\u03b82<\/sub> – \u03b81<\/sub>) …………. (2)<\/li>\n
    3. is proportional to the time of flow.
      \nQ \u221d t ………. (3) and<\/li>\n
    4. is inversely proportional to the length of the conductor.
      \nQ \u221d \\(\\frac{1}{l}\\) …………. (4)<\/li>\n<\/ol>\n

      \"TS
      \nwhere k = constant called coefficient of thermal conductivity.<\/p>\n

      Coefficient of thermal conductivity (k) :
      \nIt is defined as the amount of heat conducted normally per sec per unit area of cross-section per unit temperature gradient.
      \nS.I. Unit w m k-1<\/sup>
      \nDimensional formula = [ M\u00b9L\u00b9T-3<\/sup>\u03b8-1<\/sup>]<\/p>\n

      Problem:
      \nThermal conductivity of copper,
      \nKc<\/sub> = 401 W\/m-k
      \nTemperature at one end, \u03b82<\/sub> = 104\u00b0C
      \nTemperature of 2nd end, \u03b81<\/sub> = 24\u00b0C
      \nLength of copper bar, l = 0.1 m; Area,
      \nA = 1.0 \u00d7 10-6<\/sup> m-2<\/sup>
      \nRate of conduction,
      \n\"TS<\/p>\n

      \"TS<\/p>\n

      Question 2.
      \nState the explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable.
      \nA body cools down from 60\u00b0C to 50\u00b0C in 5 minutes and to 40\u00b0C in another 8 minutes. Find the temperature of the surroundings. [TS May ’17, ’16; AP May ’13]
      \nAnswer:
      \nNewtons’ Law of cooling :
      \nThe rate of loss of heat of the body is directly proportional to the difference of temperature of the body and the surroundings.<\/p>\n

      Explanation :
      \nThe law holds good only for small difference of temperature. Also, the loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface. We can write
      \n– \\(\\frac{dQ}{dt}\\) = k (T2<\/sub> – T1<\/sub>) (sign indicates loss) …….. (1)<\/p>\n

      where k is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass ‘m’ and specific heat capacity ‘s’ is at temperature T2<\/sub>. Let T1<\/sub> be the temperature of the surroundings. If the temperature falls by a small amount dT2<\/sub> in time dt, then the amount of heat lost is
      \ndQ = ms dT2<\/sub>
      \n\u2234 Rate of loss of hfeat is given by
      \n\"TS
      \nwhere K = k\/ms<\/p>\n

      Conditions (under which Newton’s law of cooling is applicable):
      \nNewton’s law of cooling is applicable<\/p>\n

        \n
      1. loss of heat is negligible by conduction and only when it is due to convection.<\/li>\n
      2. loss of heat occurs in a streamlined flow of air i.e., forced convection.<\/li>\n
      3. temperature of the body is uniformly distributed over it.<\/li>\n
      4. temperature difference between the body and surroundings is moderate i.e., upto 30 K.<\/li>\n<\/ol>\n

        PROBLEM :
        \nLet ‘\u03b8o<\/sub>‘ be the temperature of the surroundings.<\/p>\n

        In first case :
        \nInitial temperature, \u03b81<\/sub> = 60\u00b0C
        \nFinal temperature, \u03b82<\/sub> = 50\u00b0C
        \nTime of cooling, t = 5 minutes = 5 \u00d7 60 = 300s
        \nFrom Newton’s law of cooling we can write,
        \n\"\"<\/p>\n

        In secoend case :
        \nInitial temperature, \u03b81<\/sub> = 60\u00b0C
        \nFinal temperature, \u03b82<\/sub> = 40\u00b0C
        \nTime of cooling, t = 13 minutes = 13 \u00d7 60 = 780s
        \nAgain from Newton’s law of cooling we can write,
        \n\"TS
        \nOn solving equations (1) and (2) we get, \u03b8o<\/sub> = 33.33\u00b0C<\/p>\n

        Problems<\/span><\/p>\n

        Question 1.
        \nWhat is the temperature for which the readings on Kelvin Fahrenheit scales are same?
        \nSolution:
        \nOn the Kelvin and Fahrenheit scales
        \n\"TS
        \n\"TS<\/p>\n

        Question 2.
        \nFind the increase in temperature of aluminium rod if it’s length is to be increased by 1%. (\u03b1 for aluminium = 25 \u00d7 10-6<\/sup>\/\u00b0C) [AP Mar. ’15; June ’15]
        \nSolution:
        \nCoefficient of linear expansion of aluminium, \u03b1 = 25 \u00d7 10-6<\/sup>\/\u00b0C
        \nWe know that percentage increase in length
        \n\"TS<\/p>\n

        Question 3.
        \nHow much steam at 100\u00b0C is to be passed into water of mass 100g at 20\u00b0C to raise its temperature by 5\u00b0C? (Latent heat of steam is 540 cal \/ g and specific heat of water is 1 cal \/ g\u00b0C)
        \nSolution:
        \nLatent heat of steam, Ls<\/sub> = 540 cal\/g
        \nSpecific heat of water, Lw<\/sub> = 1 cal \/ g\u00b0C
        \nMass of water, mw<\/sub> = 100g<\/p>\n

        According to method of mixture or from the principle of calorimetry we can write, Heat lost by steam = heat gained by water
        \ni.e., ms<\/sub>Ls<\/sub> + ms<\/sub>Sw<\/sub>(100 – t) = mw<\/sub>Sw<\/sub> (t – 20)
        \n\u2234 ms<\/sub> \u00d7 540 + ms<\/sub> \u00d7 1(100 – 25)
        \n\u21d2 100 \u00d7 1 \u00d7 (25 – 20)
        \n\u21d2 615ms<\/sub> = 500(or)ms<\/sub> = \\(\\frac{500}{615}\\) = 0.813 g<\/p>\n

        \"TS<\/p>\n

        Question 4.
        \n2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ\/kg K, find the amount of heat absorbed in kJ and kcal. (J = 4.2 kJ\/kcal.)
        \nSolution:
        \nMass of air, m = 2 kg
        \nChange in temperature, \u2206T = 313 – 293 = 20K.
        \nSpecific heat at constant volume, Cv<\/sub> = 0.718 k.J\/kg – K.
        \nHeat mechanical equivalent, J = 4.2 kJ\/k.cal.
        \n\"TS
        \n\u2234 Heat energy absorbed,
        \nQ = 2 \u00d7 0.718 \u00d7 10\u00b3 \u00d7 20. = 28.72 kJ
        \n= 6.838 k calories.<\/p>\n

        Question 5.
        \nA clock, with a brass pendulum, keeps correct time at 20\u00b0C, but loses 8.212 s per day, when the temperature rises to 30\u00b0C. Calculate the coefficient of linear expansion of brass.
        \nSolution:
        \nTemperature of correct time, t1<\/sub> = 20\u00b0C
        \nLoss or gain of time in seconds per day = 8.212 sec.
        \nFinal temperature, t2<\/sub> = 30\u00b0C
        \n\u2234 \u2206t = 30 – 20 = 10
        \n\u03b1 of pendulum material = ?<\/p>\n

        In pendulum loss or gain of time in seconds per day = 43,200. \u03b1 \u2206t
        \n\"TS<\/p>\n

        Question 6.
        \nA body cools from 60\u00b0C to 40\u00b0C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10\u00b0C? [AP May ’13]
        \nSolution:
        \nIn first case :
        \nInitial temperature, \u03b81<\/sub> = 60\u00b0C
        \nFinal temperature, \u03b82<\/sub> = 40\u00b0C
        \nTime of cooling, t1<\/sub> = 7 minutes
        \n= 7 \u00d7 60 = 420s
        \nTemperature of surroundings, \u03b80<\/sub> =10\u00b0C
        \nFrom. Newton’s law of cooling, we can write,
        \n\"TS<\/p>\n

        In second case:
        \nInitial temperature, \u03b81<\/sub> = 40\u00b0C
        \nTime of cooling, t2<\/sub> = 7 minutes = 420s
        \nAgain, from Newton’s law of cooling we can write,
        \n\"TS
        \non solving equations (1) & (2) we get, 0 = 28\u00b0C<\/p>\n

        \"TS<\/p>\n

        Question 7.
        \nIf the maximum intensity of radiation for a black body is found at 2.65 \u00b5 m, what is the temperature of the radiating body? (Wein’s constant = 2.9 \u00d7 10-3<\/sup> mK)
        \nSolution:
        \nWavelength corresponding to maximum intensity, \u03bbmax<\/sub> = 2.65 \u00b5m = 2.65 \u00d7 10-6<\/sup> m.
        \nWein’s constant, b = 2.90 \u00d7 10-3<\/sup> mK.
        \nFrom Wein s Law, T = \\(\\frac{\\mathrm{b}}{\\lambda_{\\mathrm{m}}}=\\frac{2.90 \\times 10^{-3}}{2.65 \\times 10^{-6}}\\)
        \n= 1094 K.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers. TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Very Short Answer Type Questions Question 1. Distinguish between heat and temperature. [TS Mar. ’15] Answer: Differences between heat and temperature : Heat Temperature … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-study-material-chapter-12\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Telangana TSBIE\u00a0TS Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers. 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