35 - Heat and Thermodynamics Questions Answers
Three discs A,B and C having radii 2m, 4m and 6m respectively are coated with carbon black on their outer surfaces. the wavelenghts corresponding to maximum intensity are 300 nm, 400nm and 500 nm respectively. if the power radiated by them are Qa, Qb and Qc respectively then
(1) Qa is maximum
(2) Qb is maximum
(3) Qc is maximum
(4) Qa= Qb = Qc
according to Wein's displacement law λT = constant
so ratio of temperatures of the three discs will be 1/3:1/4:1/5 = 20:15:12
and ratio of area of the discs = 1:4:9
now according to Stephan's law Q α A * T4
so ratio Qa:Qb:Qc = 1*(20)4 : 4*(15)4 : 9*(12)4
on solving we will get that Qb is maximum
A bullet of mass 1 gm moving with a speed of 20m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets tuck in it. the amount of ice that melts, if 50% of the lost kinetic energy goes to ice will be
(1) 0.030 g
(2) 0.30 g
(3) 0.0003 g
(4) 3.0 g
amount of energy given by bullet = 0.5 (1/2) m v2 / 4.2 cal.
and amount of energy received by the ice block = ML here L is latent heat and M is mass of ice melted
on comparing we will get m v2/(4*4.2) = M*80 or M = 1*400/16.8*80 = 5/16.8 = 0.30 gm
a metallic sphere having inner radius a and outer radius b has thermal conductivity k = ko(k not)/r² (a ≤ r ≤ b). the thermal resistance between inner and outer surface for radiant heat flow is
consider a sphere of radius r and thickness dr
formula used is H = KAdT/L so dT/H = L/KA or R = L/KA, dT/H is thermal resistance
thermal resistance of layer of thickness dr is dr/K4πr2, on putting the value of K we get
dR = r2dr/4πr2k0
or dR = dr/4πk0
now integrate within the proper limits
Assuming the sun to be aspherical body of radius R at a temperature of T K, evaluate the total radiant power incident on earth, at a distance r from the sun. Take radius of earth as Ro(R not)
Total radiant power given by the sun = σAT4 = 4πR2σT4
amount received on the surface of earth = (πR02/4πr2)*4πR2σT4
A body emits radiant energy 1600Js‾¹ when it is at temperature 273 °C. if its temperature decreases to 273 K them it emits radiant energy at the rate of
(1) 0
(2) 800 Js‾¹
(3) 400 Js‾¹
(4) 100 Js‾¹
Temprature of second case is half of the temprature of first case so according to stefan's law radiant power will become 1/16 part = 100
A copper cube of each side 40 cm floats on mercury. the density of copper cube is 3.2g/cc and of mercury is 13.6g/cc at 27° C. the coefficient of volume expansion of Hg and linear expansion of copper are 1.8 * 10^(-4) /°C and 3 * 10^(-6) /°C respectively. Calculate the increase in height i.e. how much block sink further when the temperature rises from 27°C to 100°C?
dipped length of the cube in first case = m/Ad here m, A, d are mass of cube, area of the cube and density of mercury respectively
after increasing temperature dipped length in the second case = m/A(1+2αt)[d/(1+γt)]
values are m = VD = 403 3.2, A = 402, α = 3 * 10-6, d = 13.6, γ = 1.8 * 10-4, t = 73
all values are in CGS
now solve then substract 1st length from 2nd length
A thermally insulated piece of metal is heated by supplying a constant power P. Due to this, the temperature of the metal starts varying with time as T= at^1/4 + To(T not)
The heat capacity of the metal as a function of temperature is
(1) 4PT³/a^4
(2) 4P(T-To)³/a^4
(3) 4PT²/a³
(4) 4P(T - To)²/a³
P = dQ/dt = mcdT/dt
or mcdT = Pdt
or mc = Pdt/dT
or mc = Pd/dT[(T-T0)4/a4]
here mc is heat capacity
now solve
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between Ta and Tb?
(1) Ta = Tb/2
(2) Ta = 7Tb/4
(3) Ta = 2Tb
(4) Ta = 4Tb/7
273.15 K = 200 A
and 273.15 K = 350 B
so A = 273.15 K/200 and B = 273.15 K/350
so Ta A = TbB
or Ta/Tb = B/A = 200/350 = 4/7
so 4th option
The absolute zero temperature is
(1) -273° C
(2) -273.15 K
(3) -273.15° F
(4) -273.15° C
it is -273.15 0C
2 moles of a diatomic gas are enclosed in a vessel. when a certain amount of heat is supplied, 50% of the gas molecules get dissociated, bit there is no rise in temperature. What is the heat supplied if the temperature is T?
(1) RT
(2) RT/2
(3) 11RT/2
(4) 5RT
internal energy of 2 mole diatomic gas = 2*5/2 RT = 5RT
now after 50% disociation in monoatomic form total internal energy
of 1 mole diatomic gas = 1*5/2 RT = 2.5 RT
and that of 2 mole monoatomic = 2*3/2 RT = 3RT ( 2 mole monoatomic gas will be formed by the dissociation of 1 mole diatomic gas)
so internal energy change = 1/2 RT
This internal energy increment will be equal to the heat provided