All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions from Thermodynamics

The following questions were included from thermodynamics in the All India Pre-Medical/Pre-Dental 2009 Entrance Examination (Preliminary):

1. The internal energy change in a system that has absorbed 2 Kcals of heat and done 500 J of work is

(1) 6400 J

(2) 5400 J

(3) 7900 J

(4) 8900 J

We have ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆U is the increase in internal energy of the system and ∆W is the work done by the system.

Therefore, the increase in internal energy of the system is given by

∆U = ∆Q – ∆W = 2000×4.2 J – 500 J = (8400 – 500) J = 7900 J

[Note that 1 calorie = 4.2 joule, nearly]

2. In thermodynamic process which of the following statements is not true?

(1) In an isochoric process pressure remains constant.

(2) In an isothermal process temperature remains constant.

(3) In an adiabatic process PV^γ = constant.

(4) In an adiabatic process the system is insulated from the surroundings.

Statement (1) is not true since an isochoric process is one in which volume remains constant. So option (1) is the answer.

You will find many useful questions on thermodynamics at physicsplus here and at other locations.

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Kerala Medical Entrance Test 2006- Two Questions on Heat

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.