# 33 - Heat and Thermodynamics Questions Answers

6 identical rods are arranged in such a way that 3 of them make an equilateral triangle and other 3 are extended at the vertices forming a Y shape at each vertex. Name the triangle ABC ,  and other ends of rods as P, Q , R . given that temperatures at P , Q and R are 200°c,20°c and 20°c , what are temperatures at A , B , and C?

Joshi sir comment

A gas expands against a variable external pressure given by p = 10/V atm, where V is the volume at each stage of expansion. in expanding from 10 L to 100 L, the gas undergoes a change in internal energy ΔU = 418 J. How much heat has been absorbed?

Joshi sir comment

process is isothermal because pV = 10 = nRT

so work done W = 2.303nRTlog(V2/V1)

so Q = Δ+ W

The temperature of a perfect black body

is 1000k and its area

is 0.1m2 . the heat radiated by it in 1 minute in joule is

Joshi sir comment

Q/At = σT4

now solve

A CARNOT ENGINE WORKING B/W 270 C & --1230C HAS EFFICIENCY η . IF TEMP OF SINK IS DECREASED BY 50K THE EFFICIENCY BECOMES??

ANS  4η/3

Joshi sir comment

according to the given condition η = 1-(150/300) = 0.5

now after decreasing temperature of sink by 50K

we get the new efficiency = 1-(100/300) = 2/3 = 4*0.5/3 = 4η/3

in a thermodynamic process on 2 moles of a monoatomic gas work done on gas is 100J & change in temp is 30 C . The change in internal energy of the gas in this process is??

Joshi sir comment

change in internal energy = nCvdT = 2*[R/(γ-1)]*30 = 60*8.31/[(5/3)-1]

2 MOLES OF MONOATOMIC GAS IS MIXED WITH 1 MOLE OF A DIATOMIC  GAS. THEN γ  FOR THE MIXTURE??

ANS 1.55

Joshi sir comment

total no of freedom for this mixture = (2*3+1*5)/(2+1) = 11/3

so γ = 1+[2/f] = 1 + 2/[11/3] = 1 + [6/11] = 17/11 = 1.55

a thin square plate with each side equal to 10cm , is heated by a blacksmith . the rate radiated energy by the heated plate is 1134W the temp of hot square plate is

( σ = 5.67 X 10-8 Wm2/k4 emissivity of plate =1)

ANS 1000K

Joshi sir comment

according to the Stefans law P = AeσT4

so 1134 = 0.02*1*5.67*10-8*T4   (area of both the faces are considered)

so 1134*1010/11.34 = T4

so T = 1000

IN THE PREVIOUS QUES P2V3 =K BUT SORRY I AM NOT GETTING HOW PV3/2 =K

Joshi sir comment

take square root on both the sides

DURING AN ADIABATIC PROCESS IF THE PRESSURE OF AN IDEAL GAS IS PROPTIONAL TO THE CUBE OF ITS ABSOLUTE TEMP. THEN RATIO OF SPECIFIC HEATS AT CONSTANT PRESSURE & AT CONST. VOL IS?

ANS 3/2

Joshi sir comment

P  α   T3

so P α  (PV)3

so P2V3 = K

or PV3/2 = K

so on the basis of PVγ  =  K  calculate  γ