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| 1. |
State Newton's law of cooling verify with an experiment. |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :(i) It states that the rate of cooling of a body is directly proportional to the temperature differ between the body and the surroundings. <br/> (ii) Consider a spherical calorimeter of mass m whose outer surface is blackened. It is filled with <a href="https://interviewquestions.tuteehub.com/tag/hot-1029680" style="font-weight:bold;" target="_blank" title="Click to know more about HOT">HOT</a> water of mass m, the calorimeter with thermometer is suspended from a stand. <br/> (iii) The calorimeter & the hot water radiate heat energy to the surrounding. Using a stop clock, the temperature is noted for every 30 second interval of time see the temperature falls by about `20^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`. The readings are tabulated. <br/> If the temperature falls from `T_(1)` to `T_(2)` in t sec, the quantity of heat energy lost by radiation `<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>=(ms+m_(1)s_(1))(T_(1)-T_(2))`, where 'S' is the specific heat capacity of the <a href="https://interviewquestions.tuteehub.com/tag/material-1089170" style="font-weight:bold;" target="_blank" title="Click to know more about MATERIAL">MATERIAL</a> of the calorimeter & `S_(1)` - specific heat capacity of water. <br/> Rate of cooling = `("Heat energy lost")/("time "t_(n))` <br/> `therefore (Q)/(E)=((ms+m_(1)s_(1))(T_(1)-T_(2)))/(t)` <br/> Room temperature - `T_(0)` <br/> (v) Average excess temperature of the colorimeter over that of the surroundings = `(T_(1)+T_(2))/(2)-T_(0)` <br/> (vi) Acceleration to Newton's law of cooling `(Q)/(t)prop((T_(1)+T_(2))/(2)-T_(0))` <br/> (vii) Assume the pressure of the gas remains constant during an infinitesimally small outward displacement dy then work done `dW-F.dx=P.A.dx` <br/> `dW=P.dV` <br/> (viii) Total work done by the gas from volume `V_(1)` to `V_(2)` is <br/> `W=underset(V_(1))overset(V_(2))intP.dv` <br/> (ix) But `PV^(gamma)` = constant (k) <br/> `gamma=(C_(P))/(C_(V))` <br/> `therefore W=underset(V_(1))overset(V_(2))intkV^(gamma)dV=k[(V^(1-gamma))/(1-gamma)]_(v_(1))^(v_(2))" "[becauseP=(k)/(V^(gamma))]` <br/> `thereforeW=(k)/(1-gamma)[V_(2)^(1-gamma)-V_(1)^(1-gamma)]` <br/> `W=(1)/(1-gamma)[kV_(2)^(1-gamma)-kV_(1)^(1-gamma)]` <br/> `P_(2)V_(2)^(gamma)=P_(1)V_(2)^(gamma)=k` <br/> (x) Subtract the value of k <br/> `therefore W=(1)/(1-gamma)[P_(2)V_(2)^(gamma),V_(2)^(1-gamma)-P_(1)V_(1)^(gamma),V_(1)^(1-gamma)]` <br/> `W=(1)/(1-gamma)[P_(2)V_(2)-P_(1)V_(1)]` <br/> It `T_(2)` is the final temperature of the gas in adiable expansion, then <br/> `P_(1)V_(1)=RT_(1),P_(2)V_(2)=RT_(2)` <br/> `therefore W=(1)/(1-gamma)[RT_(2)-RT_(1)]` <br/> This is the equation for the work done during adiabating process.</body></html> | |