# 14 - Heat Transfer 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

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 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

the density of two substances are in ratio 5:6 & specific heat are in ratio 3:5. their thermal capacity per unit volume will be in ratio?

ans 1:2

Joshi sir comment

thermal capacity = ms = Vds

so thermal capacity per unit volume = ds

it takes 6 hours for ice layer thickness to grow from 1cm to 2cm on a lake in Siberia, ideally the time taken for thickness to grow from 3cm to 4cm?

ans 14 hrs

Joshi sir comment

for this type of question relation between time and thickness will be

[x2] α t

so 22 - 12 = k6

and 42 - 3= kt

on dividing we will get 3/7 = 6/t

or t = 14

Q  HEAT & WORK ARE EQUIVALENT. THIS MEANS

TEMP OF BODY CAN BE INCREASED BY DOING WORK ON IT. PLS TELL HOW THIS STAT. IS CORRECT.Q

Q IF H IS SUPPLIED TO IDEAL GAS IN ISO PROCESS. THEN W= +ve HOW? BECAUSE IF ISO THEN U= 0 & EQN IS Q=W+0 & H IS SUPPLIED SO Q=-veSO W =-ve.so ans shd be -veBUT ANS GIVEN IS +ve PLS TELL ANS WITH EXPL. THANKS FOR PRE SOLUTIONS .

Joshi sir comment

This means heat can be used to make work and work done produces heat

By doing 4.2 joule work 1 cal. heat will be produced in a body so due to this 1 cal, temp. will increase

In physics heat is supplied means Q will be +ve so W will also be positive hence is correct. A system is placed in refrigrator for absorbing heat from the body, this process will have Q -ve.

a hollow sphere of mass M (in K.G.) radius R (in m.)  is rotating with angular  frequency  omega(in radsec)  . it suddenly stops rotating and 75% of  the K.E. converted  into heat energy .   if S joulekg-kelvin is the spesific heat of material of the sphere, the rise of temperature of the sphere is ?

Joshi sir comment

Rotational K. E. = 1/2 I ω2

so 0.75 [1/2 I ω2] = M S dT

and I = 2/3 M R2

now solve

two identical conducting rods are first connected independently to  two vessels, one containing water at 100°C and the other containing ice at 0°C. In the second case, the rods are joined end to end connected to the same vessels. Let M1 and M2 g/s be the rate of melting of ice in the two cases respectively, the ratio M1/M2 is

(1) 1:2

(2) 2:1

(3) 4:1

(4) 1:4

Joshi sir comment

in first case rods are taken in parellel so net length and area will be l and 2A with conductivity coefficient K

similarly in second case rods are taken in series so net length and area will be 2l and A with conductivity coefficient K

now use H = KA(θ1- θ2)/L and  H = dQ/dt = mL/t     L is latent heat which is fixed in both the cases

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

Joshi sir comment

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 Qis 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