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Question 1. The reading tA and tB of two centigrade scale thermometers A and B graded at ice point 0°C and steam point 100°C are related by the equation tA=l +mtB+n(tB)², where l, m,n are constants. When both are immersed in oil then thermometer A reads 51°C and thermometer B reads 50 °C. determine the reading on thermometer A when thermometer B reads 25°C? Comment on the question "which thermometer is correct".

Thermodynamics Question solution, Chapter1 Zeroth law of thermodynamics. Question 1. The reading tA and tB of two centigrade scale thermometers A and B graded at ice point 0°C and steam point 100°C are related by the equation tA=l +mtB+n(tB)², where l, m,n are constants. When both are immersed in oil then thermometer A reads 51°C and thermometer B reads 50 °C. determine the reading on thermometer A when thermometer B reads 25°C? Comment on the question "which thermometer is correct". Ans- tA=l+mtB+n(tB)² According to question tA=tB=0°C tA=l+mtB+n(tB)² 0=l+0+0 l=0 Second step tA=tB=100°C tA=l+mtB+n(tB)² 100=0+m×100+n(100)² m+100n=1             ----------------------------(1) For tA=51°C tB=50°C tA=l+mtB+n(tB)² 51=0+M50+n(50)² -----------------------------(2) Now using equation (1) and (2) find m&n 50m+2500n=51 50m+5000n=50 -        -             - ________________ 0      -2500n=1                  n=-1/2500 Put value of 'n' in equati
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Question2. A new temperature scale N is to be defined. The boiling and freezing points of water on this scale are 100°N amd 400°N respectively. Then (a) what will be the reading on the new scale corresponding to 60°C? (b) At what temperature both the Celsius and new temperature scale reading would be same?

 THERMODYNAMICS  Chapter 1. ZEROTH LAW OF THERMODYNAMICS Question2. A new temperature scale N is to be defined. The boiling and freezing points of water on this scale are 100°N amd 400°N respectively. Then (a) what will be the reading on the new scale corresponding to 60°C? (b) At what temperature both the Celsius and new temperature scale reading would be same? Sol: ts= boiling temperature  ti= ice temperature  tc=corresponding temperature   scale reading:         boiling temperature    corresponding          freezing temperature  new scale                             100°N                       x°N                            400°N Celsius scale                         100°C                       60°C                           0°C Celsius scale = new scale ts - ti/ts - tc     =  ts - ti/ts - tc   100 - 0/100 - 60 = 100°N - 400°N/100°N - X°C 100/40 = -300°N/100°N - X°N 100°N - X°N = -120°N    X = 220°N (b)  both Celsius and new temperature scale value same at 100°
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Question 1. A spherical balloon of 1m diameter contains a gas at 150kpa. The gas inside balloon is heated until pressure reaches 459kpa. During the process of heating the pressure of gas inside the balloon is proportional to cube of the diameter of the balloon. Find the work done by the gas.

THERMODYNAMICS QUESTION SOLUTION, CHAPTER. ENERGY INTERACTION. Question 1. A spherical balloon of 1m diameter contains a gas at 150 KPa. The gas inside balloon is heated until pressure reaches 450 KPa. During the process of heating the pressure of gas inside the balloon is proportional to cube of the diameter of the balloon. Find the work done by the gas. Solution: D1= 1 m    P1= 150 KPa                                           V = (4/3)  π R^3 =    πD^3/6        P2 = 450 KPa                                             dV = ( π/6)  (3D^2)    δD P directly proportional D^3                         dV = ( π/2) (D^2)   δD P = k(D^3)                                                   W =  ∫ P dV  --------------(0) P1=k(D1^3)                                                  150 = k(1)^3                                              k = 150 KN/m^3 ----------------(1)                P2 = k(D2^3) 450 = 150(D2)^3 D2 = 1.44 m -----------------------(2) W =  ∫ P dV  W =  ∫ (kD^3)   ( π/2) (D^2)   δ
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