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A piece of metal is heated to a temperature theta and then left to cool in a room maintained at temperature theta0 (with theta0 less than theta). Which of the following best describes the graph of the metal's temperature T versus time t?
- A curve that starts at theta, falls steeply at first and then more gradually, approaching the room temperature theta0 asymptotically (exponential decay).
- A straight line falling at a constant rate from theta down to theta0.
- A curve that falls slowly at first and then more and more steeply toward theta0.
- A horizontal straight line that stays constant at theta for all time.
Correct answer: A curve that starts at theta, falls steeply at first and then more gradually, approaching the room temperature theta0 asymptotically (exponential decay).
Solution
By Newton's law of cooling, dT/dt = -k(T - theta0), whose solution is T = theta0 + (theta - theta0) e^(-k t). This is an exponential decay from theta toward theta0: steepest near t = 0 (large T - theta0) and flattening as it approaches the asymptote theta0. It is not linear and never falls below theta0.
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
Group-I:
(P) 2000 K
(Q) 3000 K
(R) 5000 K
(S) 10000 K
Group-II:
(1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV.
(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
(4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K.
(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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