1Exercises20 questions
Q.10.1The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.v
SolutionUse $t_C=T-273.15$ and $t_F=\frac95t_C+32$. For neon, $t_C=24.57-273.15=-248.58^\circ\text{C}$ and $t_F=-415.44^\circ\text{F}$. For carbon dioxide, $t_C=216.55-273.15=-56.60^\circ\text{C}$ and $t_F=-69.88^\circ\text{F}$.
Answer:Neon: $-248.58^\circ\text{C}$, $-415.44^\circ\text{F}$. Carbon dioxide: $-56.60^\circ\text{C}$, $-69.88^\circ\text{F}$.
Q.10.2Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ?v
SolutionBoth are absolute scales, so zero is absolute zero. For the same temperature, scale readings are proportional to their triple-point readings: $T_A/200=T_B/350$. Hence $T_A=(200/350)T_B=\frac47T_B$.
Answer:$T_A=\frac{4}{7}T_B$.
Q.10.3The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = Ro [1 + α (T – To )]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?v
SolutionTake $R_0=101.6\,\Omega$ at $T_0=273.16\,\text{K}$. From the lead point, $\alpha=[165.5/101.6-1]/(600.5-273.16)=1.921\times10^{-3}\,\text{K}^{-1}$. For $R=123.4\,\Omega$, $T=T_0+[R/R_0-1]/\alpha=273.16+[123.4/101.6-1]/\alpha=384.8\,\text{K}$.
Answer:$384.8\,\text{K}$.
Q.10.4Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. 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 Celsius scale as mentioned above which were assigned the number 0 °C and 100 °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 K. What is the other fixed point on this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) 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 ?v
SolutionThe triple point fixes a unique state where ice, water and vapour coexist. Ice point and steam point vary with pressure and purity. The other natural fixed point of an absolute scale is absolute zero. Since the triple point is $273.16\,\text{K}$ and $0.01^\circ\text{C}$, the Celsius-Kelvin offset is $273.15$. A Fahrenheit degree is $5/9$ of a kelvin, so the triple-point reading on an absolute scale with Fahrenheit-size intervals is $273.16\times9/5=491.69$.
Answer:(a) The triple point is unique and reproducible; melting and boiling points depend on pressure and impurities. (b) Absolute zero, $0\,\text{K}$. (c) The triple point of water is $0.01^\circ\text{C}$, so $0^\circ\text{C}=273.15\,\text{K}$. (d) $491.69$ on that absolute Fahrenheit-size scale.
Q.10.5Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :
Temperature
Pressure thermometer A
Pressure thermometer B
Triple-point of water
1.250 × 105 Pa
0.200 × 105 Pa
Normal melting point of sulphur
1.797 × 105 Pa
0.287 × 105 Pa
(a) What is the absolute temperature of 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 ?v
SolutionAt constant volume, $T=273.16(P/P_{tp})$. For A, $T=273.16(1.797/1.250)=392.7\,\text{K}$. For B, $T=273.16(0.287/0.200)=392.0\,\text{K}$. The slight difference arises from real-gas deviations from ideal behaviour. Repeating at lower pressures and extrapolating to zero pressure gives a common ideal-gas value.
Answer:(a) Thermometer A: $392.7\,\text{K}$; thermometer B: $392.0\,\text{K}$. (b) The gases are not perfectly ideal at finite pressure; reducing pressure and extrapolating to zero pressure reduces the discrepancy.
Q.10.6A 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 .v
SolutionAt $45^\circ\text{C}$, each tape division is larger by factor $1+\alpha\Delta T=1+1.20\times10^{-5}(18)$. Thus actual length $=63.0[1+1.20\times10^{-5}(18)]=63.0136\,\text{cm}$. Since rod and tape are both steel, cooling both to the calibration temperature removes the same expansion factor, giving $63.0\,\text{cm}$.
Answer:Actual length at $45.0^\circ\text{C}$ is $63.0136\,\text{cm}$. Its length at $27.0^\circ\text{C}$ is $63.0\,\text{cm}$.
Q.10.7A 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 coefficient of linear expansion of the steel to be constant over the required temperature range :
αsteel = 1.20 × 10–5 K–1.v
SolutionThe shaft must shrink from $8.70\,\text{cm}$ to $8.69\,\text{cm}$. Use $D=D_0[1+\alpha(T-27)]$. Thus $8.69=8.70[1+1.20\times10^{-5}(T-27)]$. Solving gives $T=-68.8^\circ\text{C}$.
Answer:About $-69^\circ\text{C}$.
Q.10.8A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.v
SolutionA hole in a sheet expands as if it were made of the same material. $\Delta D=D\alpha\Delta T=4.24(1.70\times10^{-5})(200)=0.0144\,\text{cm}$.
Answer:The diameter increases by $0.0144\,\text{cm}$, or $0.144\,\text{mm}$.
Q.10.9A 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.v
SolutionSince supports are rigid, prevented thermal strain $\alpha\Delta T$ becomes tensile strain. Here $\Delta T=66\,\text{K}$ and area $A=\pi(1.0\times10^{-3})^2=3.14\times10^{-6}\,\text{m}^2$. Tension $F=YA\alpha\Delta T=(0.91\times10^{11})(3.14\times10^{-6})(2.0\times10^{-5})(66)=3.77\times10^2\,\text{N}$.
Answer:$3.8\times10^2\,\text{N}$ approximately.
Q.10.10A 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 ).v
SolutionTemperature rise is $210\,\text{K}$. Brass expansion: $\Delta L_b=0.50(2.0\times10^{-5})(210)=0.00210\,\text{m}=0.210\,\text{cm}$. Steel expansion: $\Delta L_s=0.50(1.2\times10^{-5})(210)=0.00126\,\text{m}=0.126\,\text{cm}$. Total change $=0.336\,\text{cm}$. Since the joined rod is free to expand, no thermal stress is set up.
Answer:Total increase in length is $0.336\,\text{cm}$. No thermal stress is developed because the ends are free to expand.
Q.10.11The coefficient of volume expansion of glycerine is 49 × 10–5 K–1. What is the fractional change in its density for a 30 °C rise in temperature ?v
SolutionFor fixed mass, density is inversely proportional to volume. For small expansion, $\Delta\rho/\rho\approx-\Delta V/V=-\gamma\Delta T=-(49\times10^{-5})(30)=-1.47\times10^{-2}$.
Answer:Density decreases by about $1.47\times10^{-2}$, i.e. $1.47\%$.
Q.10.12A 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.v
SolutionOnly 50% of the power heats the block, so heat supplied is $Q=0.5(10^4)(2.5\times60)=7.5\times10^5\,\text{J}$. Specific heat is $0.91\,\text{J g}^{-1}\text{K}^{-1}=910\,\text{J kg}^{-1}\text{K}^{-1}$. Thus $\Delta T=Q/(ms)=7.5\times10^5/(8.0\times910)=103\,\text{K}$.
Answer:About $103\,\text{K}$.
Q.10.13A 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 ).v
SolutionMaximum melting occurs if the copper cools to $0^\circ\text{C}$. Heat lost by copper is $Q=ms\Delta T=(2500\,\text{g})(0.39)(500)=4.875\times10^5\,\text{J}$. Ice melted $m=Q/L_f=4.875\times10^5/(335\,\text{J g}^{-1})=1455\,\text{g}=1.46\,\text{kg}$.
Answer:$1.46\,\text{kg}$ of ice.
Q.10.14In 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 ?v
SolutionWater plus calorimeter water equivalent is $0.150+0.025=0.175\,\text{kg}$. Heat gained $=0.175(4186)(40-27)$. Heat lost by metal $=0.20 c(150-40)$. Equating gives $c=0.175(4186)(13)/(0.20\times110)=433\,\text{J kg}^{-1}\text{K}^{-1}$. If heat is lost to surroundings, the metal actually lost more heat than counted, so the actual specific heat is larger.
Answer:$4.33\times10^2\,\text{J kg}^{-1}\text{K}^{-1}$, or about $0.103\,\text{cal g}^{-1}\text{K}^{-1}$. If heat losses are not negligible, this computed value is smaller than the actual value.
Q.10.15Given below are observations on molar specific heats at room temperature of some common gases.
Gas
Molar specific heat (Cv )
(cal mo1–1 K–1)
Hydrogen
4.87
Nitrogen
4.97
Oxygen
5.02
Nitric oxide
4.99
Carbon monoxide
5.01
Chlorine
6.17
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 ?v
SolutionA monatomic gas has only three translational degrees of freedom, giving $C_V=\frac32R\approx2.98\,\text{cal mol}^{-1}\text{K}^{-1}$. Diatomic gases at room temperature also have rotational degrees of freedom, giving about $C_V=\frac52R\approx4.97\,\text{cal mol}^{-1}\text{K}^{-1}$, close to most values in the table. Chlorine's larger value indicates additional vibrational contribution.
Answer:These gases are diatomic, so rotational degrees of freedom add to their heat capacity. Chlorine's larger value suggests vibrational degrees of freedom also contribute appreciably at room temperature.
Q.10.16A 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.v
SolutionTemperature drop is $3^\circ\text{F}=3\times5/9=1.67^\circ\text{C}$. Heat lost by the body is $Q=mc\Delta T=(30000\,\text{g})(1\,\text{cal g}^{-1}\text{K}^{-1})(1.67)=5.0\times10^4\,\text{cal}$. Sweat evaporated $=Q/L=5.0\times10^4/580=86.2\,\text{g}$. Over 20 min, rate $=86.2/20=4.3\,\text{g min}^{-1}$.
Answer:$4.3\,\text{g min}^{-1}$ approximately.
Q.10.17A ‘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 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]v
SolutionApproximate conducting area of the six faces as $A=6(0.30)^2=0.54\,\text{m}^2$, thickness $L=0.05\,\text{m}$, and temperature difference $45\,\text{K}$. Heat entering in 6 h is $Q=KA\Delta Tt/L=0.01(0.54)(45)(21600)/0.05=1.05\times10^5\,\text{J}$. Ice melted $=Q/L_f=1.05\times10^5/(335\times10^3)=0.313\,\text{kg}$. Remaining ice $=4.0-0.313=3.69\,\text{kg}$.
Answer:About $3.69\,\text{kg}$ of ice remains.
Q.10.18A 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.v
SolutionHeat needed to boil water at $6.0\,\text{kg/min}=0.10\,\text{kg/s}$ is $H=0.10(2256\times10^3)=2.256\times10^5\,\text{W}$. Conduction through the base gives $H=KA(T-100)/L$. Thus $T-100=HL/(KA)=(2.256\times10^5)(0.01)/(109\times0.15)=138^\circ\text{C}$, so $T\approx238^\circ\text{C}$.
Answer:About $238^\circ\text{C}$.
Q.10.19Explain 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 waterv
SolutionBy Kirchhoff's law, good absorbers are good emitters; a highly reflective body absorbs little and emits poorly. Brass has much higher thermal conductivity than wood, so it removes heat from the skin faster and feels colder. An optical pyrometer calibrated for a black body underestimates an open iron piece because emissivity is less than one; inside a furnace, cavity radiation makes it effectively black. The atmosphere provides greenhouse warming. Steam heating is efficient because condensing steam releases latent heat in addition to cooling heat.
Answer:(a) Good reflectors are poor absorbers and hence poor emitters. (b) Brass conducts heat away from the hand faster than wood. (c) Red-hot iron in the open is not a black body, but in a furnace it behaves nearly like one. (d) The atmosphere reduces heat loss by absorbing/re-emitting infrared radiation. (e) Steam releases large latent heat when it condenses.
Q.10.20A 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.v
SolutionUsing Newton's cooling law in integrated form, $T-T_s$ decays exponentially. For the first interval, $(80-20)/(50-20)=60/30=2=e^{5K}$. For cooling from $60^\circ\text{C}$ to $30^\circ\text{C}$, $(60-20)/(30-20)=40/10=4=e^{Kt}$. Since $4=2^2$, $t=10\,\text{min}$.