Miscellaneous Exercises Thermal Physics:-

Problem 1:-Icebergs in the North Atlantic present hazards to shipping as shown in the below figure, causing the length of shipping routes to increase by about 30% during the iceberg season. Strategies for destroying icebergs include planting explosives, bombing, torpedoing, shelling, ramming, and painting with lampblack. Suppose that direct melting of the iceberg , by placing heat sources in the ice, is tried. How much heat is required to melt 10% of a 210,000 – metric-ton iceberg? (One metric ton = 1000 kg.)

Concept:-The amount of heat per unit mass that must be transferred to produce a phase change is called the latent

Lfor the process. The total heat transferred in a phase change is then

Q=Lm,where m is the mass of the sample that changes phase. The heat transferred during melting or freezing is called the heat of fusion.

Solution:-To obtain the heat that is required to melt 10% of a 210,000-metric-ton iceberg, substitute 210,000-metric-ton for mass of the iceberg

mand 333 ×10^{3}J/kg for latent heat of fusion of water in the equationQ= 10% (Lm),Q = 10% (Lm)

= 0.10 (333 ×10

^{3}J/kg) (210,000-metric-ton)= 0.10 (333 ×10

^{3}J/kg) (210,000-metric-ton) (10^{3}kg/1 metric-ton)= 0.10(333 ×10

^{3}J/kg) (2.1×10^{8}kg)= 7.0×10

^{12}JFrom the above observation we conclude that, the heat that required to melt 10% of a 210,000-metric-ton iceberg would be 7.0×10

^{12}J.___________________________________________________________________________

Problem 2:-For the Carnot cycle shown in below figure, calculate (a) the heat that enters and (b) the work done on the system.

Concept:-In an isothermal process, the heat

Qtransfer takes place at constant temperatureT. The entropy ΔSof the system is defined as,Δ

S=Q/THere

Qis the positive heat flow into the system. So the entropy ΔSof the system will increase.From the equation Δ

S=Q/T, the absorbed heatQwill be,Q = (ΔS) (T)

= T (S

_{f}– S_{i})Here S

_{f}is the final entropy of the system and S_{i}is the initial entropy of he system.For a cyclic path,

Q+W= 0Here

Qis the heat added to the system andWis the work done.

Solution:-(a) We have to calculate the heat

Q_{in}that enters into the system.In the above cycle, the heat only enters along the top path

AB.To obtain the heat

Q_{in}that enters into the system, substitute 400 K forT, 0.6 J/K forS_{f}and 0.1 J/K forS_{i}in the equationQ=T(S_{f}–S_{i}),

Q_{in}=T(S_{f}–S_{i})= (400 K) (0.6 J/K - 0.1 J/K)

= (400 K) (0.5 J/K)

= 200 J

Therefore the heat

Q_{in}that enters into the system would be 200 J.(b) To find the work done, first we have to find out the heat that leaves

Q_{out}from the bottom path of the cycle.In the above cycle, the heat only leaves along the bottom path

CD.

Q_{out}=T(S_{f}–S_{i})= (250 K) (0.1 J/K - 0.6 J/K)

= (250 K) (-0.5 J/K)

= -125 J

Thus the heat that leaves

Q_{out}from the bottom path of the cycle would be -125 J.As,

Q+W= 0 for a cyclic path, so work doneWwill be,

W= -Q= - (Q

_{in}+ Q_{out}) (Since, Q = Q_{in}+ Q_{out})To obtain the work done W on the system, substitute 200 J for Q

_{in}and -125 J for Q_{out}in the equation W = - (Q_{in}+ Q_{out}),W = - (Q

_{in}+ Q_{out})= - (200 J+(-125 J) )

= -75 J

From the above observation we conclude that, the work done W on the system would be -75 J.

______________________________________________________________________________

Problem 3:-A Carnot engine works between temperatures T

_{1}and T_{2}. It derives a Carnot refrigerator that works between two different temperatures T_{3}and T_{4}as shown in the below figure. Find the ratio |Q_{3}|_{ }/ |Q_{1}| in terms of the four temperatures.

Concept:-The efficiency

εof a Carnot engine is equal to the,ε = 1- T

_{L}/T_{H}= |W|/|Q

_{H}|Here, T

_{L}is the temperature of sink (lower temperature), T_{H}is the temperature of source, W is the work done by the engine and Q_{H}is the heat at higher temperature.The coefficient of performance

Kof the Carnot refrigerator will be,K = |Q

_{L}|/W= T

_{L}/T_{H}-T_{L}Here T

_{L}is lower temperature, T_{H}is the higher temperature, W is the work done by the engine and Q_{L}is the heat at lower temperature.

Solution:-To obtain the efficiency

εof the Carnot engine, substituteT_{2}forT_{L},T_{1}forT_{H}andQ_{1}forQ_{H}in the equationε= 1-T_{L}/T_{H}=|W|/|Q_{H}|,

ε=1-T_{2}/T_{1}= |W|/|Q

_{1}|To obtain the coefficient of performance

Kof the Carnot refrigerator, substituteT_{4}forT_{L},T_{3}forT_{H}andQ_{4}forQ_{L}in the equationK=T_{L}/T_{H}-T_{L}= |Q_{L}|/W,

K=T_{4}/T_{3}-T_{4}= |Q

_{4}|/|W|Lastly, |

Q_{4}| = |Q_{3}| - |W|.We just need to combine these three expressions into one.

From

ε=1-T_{2}/T_{1}= |W|/|Q_{1}|, we get,1-T

_{2}/T_{1}= |W|/|Q_{1}||W| = |Q

_{1}| T_{1}-T_{2}/T_{1}From

K=T_{4}/T_{3}-T_{4}=|Q_{4}|/|W|, we get,

T_{4}/T_{3}-T_{4}= |Q_{4}|/|W|= |Q

_{3}| - |W|/|W| (Since, |Q_{4}| = |Q_{3}| - |W|)= |Q

_{3}|/|W| - 1Substitute the value of |Q

_{1}| T_{1}-T_{2}/T_{1}for W in the equation T_{4}/T_{3}-T_{4}=|Q_{3}|/|W| - 1, we get,T

_{4}/T_{3}-T_{4}=|Q_{3}|/|W| - 1=|Q

_{3}|/|Q_{1}| (T_{1}/T_{1}-T_{2})- 1So, |Q

_{3}|/|Q_{1}| = (T_{4}/T_{3}-T_{4}+1) (T_{1}-T_{2}/T_{1})= (T

_{3}/T_{3}-T_{4}) (T_{1}-T_{2}/T_{1})= (1- T

_{2}/T_{1}) / (1- T_{4}/T_{3})From the above observation we conclude that, the ratio|Q

_{3}|/|Q_{1}| in terms of the four temperatures would be equal to (1- T_{2}/T_{1}) / (1- T_{4}/T_{3}).______________________________________________________________________________________

Problem 4:-One mole of an ideal monoatomic gas is caused to go through the cycle as shown in below figure. Hw much work is done on the gas in expanding the gas from a to c along path abc? (b) What is the change in internal energy and entropy in going from b to c? (c) What is the change in internal energy and entropy in going through one complete cycle? Express all answers in terms of the pressure p

_{0}and volume V_{0 }at a point a in the diagram.

Concept:-Work done

Wat constant pressure is defined as,

W=pΔVHere p is the pressure and Δ

Vis the change in volume.Change in internal energy Δ

Eis defined as,ΔE = 3/2 nRΔT

Here

nis the number of moles,Ris the gas constant and ΔTis the change in temperature.Change in entropy Δ

Sis defined as,ΔS = 3/2 ln (T

_{f}/T_{i})Here,

T_{f}is the final temperature andT_{i}is the initial temperature.

Solution:-

(a)To obtain the work doneWalong the path_{abc}abc, first we have to find out the work doneWalong the path_{ab}ab.From the figure we observed that,

W=_{ab}pΔV= p

_{0}(V_{0}-4V_{0})= -3 p

_{0}V_{0}So, W

_{abc}= W_{ab}= -3 p

_{0}V_{0}From the above observation we conclude that, the work done on the gas along the path abc would be -3 p

_{0}V_{0}.

(b)The change in internal energy ΔEalong the path_{bc}btocwould be,Δ

E= 3/2_{bc}nRΔT_{bc}= 3/2 (nRT

_{c}– nRT_{b})= 3/2(p

_{c}V_{c }– p_{b}V_{b}) (pV = nRT)= 3/2[(2p

_{0}) (4V_{0}) -(p_{0}) (4V_{0})]= 3/2 [8 p

_{0}V_{0}- 4 p_{0}V_{0}]= 6 p

_{0}V_{0}The change in entropy ΔS

_{bc}along the path b to c would be,ΔS

_{bc}= 3/2nR ln(T_{c}/T_{b})= 3/2nR ln(P

_{c}/P_{b}) (Since, T_{c}/T_{b}= P_{c}/P_{b})=3/2nR ln(2P

_{0}/P_{0})= 3/2 nR ln2

From the above observation we conclude that, the change in internal energy ΔE

_{bc}in the path b to c would be 6 p_{0}V_{0}and the change in entropy ΔS_{bc}in the path b to c would be 3/2 nR ln2.

(c)The change in entropy and internal energy in a cyclic process is zero. Since the process is a complete cycle, therefore the change in entropy and internal energy in going through one complete cycle would be zero._______________________________________________________________________________________________

Problem 5:-(a) Calculate the rate of heat loss through a glass window of area 1.4 m

^{2}and thickness 3.0 mm if the outside temperature is -20ºF and the inside temperature is +72ºF. (b) A storm window is installed having the same thickness of glass but with an air gap of 7.5 cm between the two windows. What will be the corresponding rate of heat loss presuming that conduction is the only important heat-loss mechanism?

Concept:-The rate

Hat which heat is transferred through the rod is,(a) directly proportional to the cross-sectional area (A) available.

(b) inversely proportional to the length of the rod Δx.

(c) directly proportional to the temperature difference ΔT.

So,

H=kAΔT/ ΔxHere k is the proportionality constant and is called thermal conductivity of the rod.

Again the rate of heat transfer

His defined as,

H=A(T_{2}-T_{1})/∑R_{n}= A ΔT/ (R

_{g}+ R_{a})Here

Ais the area, ΔTis the temperature difference,R_{g}is the thermal resistance of glass at the thickness 3.0 mm andR_{a}is the thermal resistance of air at the thickness 7.5 cm.The thermal resistance or

R-value, defined by,

R=L/kHere

L is the thickness of the material through which the heat is transferred andkis the thermal conductivity of the material.

Solution:-

(a)First we have to find the temperature difference between inside and outside of the glass window.If

T_{2}is the inside temperature andT_{1}is the outside temperature, then the temperature difference ΔTin^{̊}C will be,Δ

T=T_{2}-T_{1}= 5

^{̊}C /9^{̊}F (72^{̊}F –(-20^{̊ }F))= 51.1

^{̊}C= (51.1+273) K

= 324.1 K

To obtain the rate of heat loss through a glass window, substitute 1.0 W/m.K (thermal conductivity of window glass) for

k, 1.4 m^{2}for A, 324.1 K for ΔTand 3.0 mm for Δxin the equationH=kAΔT/ Δx, we get,

H=kAΔT/ Δx= (1.0 W/m.K) (1.4 m

^{2}) (324.1 K)/ (3.0 mm)= (1.0 W/m.K) (1.4 m

^{2}) (324.1 K)/ (3.0 mm×10^{-3}m/1 mm)= 2.4×10

^{4}WFrom the above observation we conclude that, the rate of heat loss through a glass window would be 2.4×10

^{4}W.

(b)First we have to find out the thermal resistanceR_{g}of galss at the thickness 3.0 mm andR_{a}of air at the thickness 7.5 cm.To obtain the thermal resistance

R_{g}of galss at the thickness 3.0 mm, substitute 3.0 mm forLand 1.0 W/m.K (thermal conductivity of window glass) forkin the equationR=L/k,

R_{g}=L/k= 3.0 mm/1.0 W/m.K

= (3.0 mm×10

^{-3}m/1 mm) /(1.0 W/m.K)= 3.0×10

^{-3}m^{2}.K/WTo obtain the thermal resistance

R_{a}of air at the thickness 7.5 cm, substitute 7.5 cm forLand 0.026 W/m.K (thermal conductivity of dry air) forkin the equationR=L/k,

R_{a}=L/k= 7.5 cm /0.026 W/m.K

= (7.5 cm ×10

^{-2}m/1 mm) /(0.026 W/m.K)= 2.88 m

^{2}.K/WTo find out the corresponding rate of heat loss

H, substitute 1.4 m^{2}forA,3.0×10^{-3}m^{2}.K/W forR_{g}and 2.88 m^{2}.K/W forR_{a}in the equationH=AΔT/ (R_{g}+R_{a}), we get,

H=AΔT/ (R_{g}+R_{a})= (1.4 m

^{2}) (324.1 K) /(3.0×10^{-3}m^{2}.K/W) (2.88 m^{2}.K/W)= 25 W

From the above observation we conclude that, the corresponding rate of heat loss

Hwould be 25 W.___________________________________________________________________________________________________

Problem 6:-Consider the slab as shown in the below figure. Suppose that Δx = 24.9 cm, A = 1.80 m

^{2}, and the material is copper. If T = -12.0ºC, ΔT = 136ºC, and a steady state is reached, find (a) the temperature gradient, (b) the rate of heat transfer, and (c) the temperature at a point in the rod 11.0 cm from the high temperature end.

Concept:-The temperature gradient is defined as,

Temperature gradient = Δ

T/ΔxHere Δ

Tis the temperature difference and Δxis the thickness of the slab.The rate

Hat which heat is transferred through the slab is,(a) directly proportional to the area (

A) available.(b) inversely proportional to the thickness of the slab Δx.

(c) directly proportional to the temperature difference Δ

T.So,

H=kAΔT/ ΔxHere k is the proportionality constant and is called thermal conductivity of the material.

Solution:-(a) To obtain the temperature gradient, substitute 136

^{̊}C for ΔTand 0.249 m for Δxin the equation, temperature gradient = ΔT/Δx,Temperature gradient = Δ

T/Δx= (136

^{̊}C)/(24.9 cm)= (136 +273) K/( 24.9 cm×10

^{-2}m/1 cm)= 409 K /0.249 m

= 1643 K/m

Therefore the temperature gradient of the slab will be 1643 K/m.

(b) To find out the rate of heat transferHof the slab, substitute 1.80 m^{2}for the areaA, 401 W/m. K for the thermal conductivitykof copper and1643 K/m for ΔT/Δxin the equationH=kAΔT/ Δx,H = kA ΔT/ Δx

= (401 W/m. K) (1.80 m

^{2}) (1643 K/m)= 11.85×10

^{5}WTherefore the rate of heat transfer H of the slab will be 11.85×10

^{5}W.(c) ΔT = T_{H}– THere

Tis the temperature at a point in the rod 11.0 cm from the high temperature end.So,

T_{H}= ΔT+TTo find out

T_{H}, substitute 136^{̊}C for ΔTand -12^{̊}C forTin the equationT_{H}= ΔT+T,

T_{H}= ΔT+T= (136 +273) K + (-12+ 273 )

= 409 K + 261 K

= 670 K

We know that temperature gradient is,

Δ

T/Δx= 1643 K/m(

T_{H}–T)/Δx= 1643 K/m (Since, ΔT=T_{H}–T)

T_{H}–T= (1643 K/m) Δx

T=T_{H}- (1643 K/m) ΔxTo find out

T, substitute 670 K forT_{H}and 11.0 cm for Δxin the equationT=T_{H}- (1643 K/m) Δx,

T=T_{H}- (1643 K/m) Δx= (670 K) – ((1643 K/m) (11.0 cm))

= (670 K) – ((1643 K/m) (11.0 cm))

= (670 K) – ((1643 K/m) (11.0 cm×10

^{-2}m/1 cm ))= (670 K) – ((1643 K/m) (0.11 m)

= (670 K) – (180.73 K)

= 489.27 K

Rounding off to two significant figures, the temperature at a point in the rod 11.0 cm from the high temperature end will be 490 K.

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