The following two questions appeared in the Kerala Medical Entrance 2006 test paper:
(1) A thermos flask made of stainless steel contains several tiny lead shots. If the flask is quickly shaken up and down several times, the temperature of lead shots
(a) increases by adiabatic process (b) increases by isothermal process (c) decreases by adiabatic process (d) remains the same (e) first decreases and then increases.
This is a simple question aimed at testing your understanding of a basic concept. In an adiabatic process, there is no heat transfer to or from the system. The system in the question is the lead shots contained in the thermos flask which is thermally insulated from outside. The temperature of the lead shots is therefore increased by adiabatic process [Option (a)].
(2) A body cools from 50˚C to 49.9˚C in 5 s. How long will it take to cool from 40˚C to 39.9˚C? Assume the temperature of the surroundings to be 30˚C and Newton’s law of cooling to be valid.
(a) 2.5 s (b) 5 s (c) 20 s (d) 10 s (e) 15 s
As per Newton’s law of cooling, the rate of cooling of a body is directly proportional to the mean excess of temperature of the body over the surroundings (if the absolute temperature of the body is nearly equal to the absolute temperature of the surroundings).
Therfore, we have (50 – 49.9)/5 α (49.95 – 30) and
(40 – 39.9)/t α (39.95 – 30)
Dividing the first one by the second, we obtain t/5 = 2, very nearly, so that the required time for cooling is 10 s.
(1) A thermos flask made of stainless steel contains several tiny lead shots. If the flask is quickly shaken up and down several times, the temperature of lead shots
(a) increases by adiabatic process (b) increases by isothermal process (c) decreases by adiabatic process (d) remains the same (e) first decreases and then increases.
This is a simple question aimed at testing your understanding of a basic concept. In an adiabatic process, there is no heat transfer to or from the system. The system in the question is the lead shots contained in the thermos flask which is thermally insulated from outside. The temperature of the lead shots is therefore increased by adiabatic process [Option (a)].
(2) A body cools from 50˚C to 49.9˚C in 5 s. How long will it take to cool from 40˚C to 39.9˚C? Assume the temperature of the surroundings to be 30˚C and Newton’s law of cooling to be valid.
(a) 2.5 s (b) 5 s (c) 20 s (d) 10 s (e) 15 s
As per Newton’s law of cooling, the rate of cooling of a body is directly proportional to the mean excess of temperature of the body over the surroundings (if the absolute temperature of the body is nearly equal to the absolute temperature of the surroundings).
Therfore, we have (50 – 49.9)/5 α (49.95 – 30) and
(40 – 39.9)/t α (39.95 – 30)
Dividing the first one by the second, we obtain t/5 = 2, very nearly, so that the required time for cooling is 10 s.
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