Kerala Engineering Entrance 2008 Questions on Heat & Thermodynamics

The following four questions on heat and thermodynamics appeared in KEAM (Engineering) 2008 question paper:

(1) Three rods made of the same material and having the same cross section have been joined as shown in the figure. Each rod is of same length. The temperature at the junction of the three rods is

(a) 45º C

(b) 90º C

(c) 30º C

(d) 20º C

(e) 60º C

The quantity of heat flowing through the two hotter rods has to flow through the colder rod so that we have

2×KA(90º – T)/L = KA(T– 0)/L where K, A and L are respectively the thermal conductivity, area of cross section and the length of each rod and T is the temperature of the junction in degree Celsius.

This gives 180º – 2T = T from which T = 60º

(2) The PV diagram of a gas undergoing a cyclic process (ABCDA) is shown in the graph where P is in units of Nm^–2 and V is in cm^–3. Identify the incorrect statement

(a) 0.4 J of work is done by the gas from A to B

(b) 0.2 J of work is done by the gas from C to D

(c) No work is done by the gas from B to C

(d) Net work is done by the gas in one cycle is 0.2 J

(e) Work is done by the gas in going from B to C and on the gas fromD to A

BC and DA represent isochoric processes (in which volume does not change) and hence no work is done. So, the incorrect statement is (e).

[Since Kerala entrance questions are single answer type there is no need of checking the other options when you are actually facing the examination. But it will be usful to know the correctness of the other options at the present moment.

Work (PdV) which is equal to 2×10⁵×2×10^–6 = 0.4 J is done by the gas from A to B since the gas expands.

Work (PdV) which is equal to 1×10⁵×2×10^–6 = 0.2 J is done on the gas from C to D since the gas gets compressed.

Since the cyclic process is clockwise, a net amount of work is done by the gas and its value is equal to the area of the closed curve which is 1×10⁵×2×10^–6 = 0.2 J].

(3) The plots of intensity of radiation versus wave length of three black bodies at temperatures T₁, T₂ and T₃ are shown. Then

(a) T₃> T₂> T₁

(b) T₁> T₂> T₃

(c) T₂> T₃> T₁

(d) T₁> T₃> T₂

(e) T₃> T₁> T₂

According to Wien’s law the product λ_mT is a constant (equal to 0.29 cmK) in the case of any black body (where λ_m and T are respectively the wave length of emitted radiation of maximum intensity and T is the temperature of the black body). Obviously, option (d) is correct.

(4) A bubble of 8 moles of helium is submeged at a certain depth in water. The temperature of water increases by 30º C. How much heat is added approximately to helium during expansion?

(a) 4000 J

(b) 3000 J

(c) 3500 J

(d) 4500 J

(e) 5000 J

The heat δQ added to helium is given by

δQ = mC_PδT wher m is the mass of helium, C_P is its specific heat at constant pressure and δT is its rise in temperature.

[You have to use C_P (and not the specific heat at constant volume C_V) since the bubble absorbs the heat at constant pressure. Further, since we are given the amount of the gas in moles, the molar specific heat is to be used in the above relation].

Therefore, δQ = 8×(5/2)R×30, noting that helium is mono atomic so that its C_P = (5/2)R where R is universal gas constant which is approximately equal to 8.3 J mol^–1K^–1

Thus δQ = 8×(5/2) ×8.3×30 = 5000 J, nearly.

You will find a useful post on the equations to be remembered in thermodynamics here