# 13 - Heat Transfer Questions Answers

The temperature of a perfect black body

is 1000k and its area

is 0.1m^{2} . the heat radiated by it in 1 minute in joule is

**Asked By: HASITHA.KOTTA**

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**Joshi sir comment**

Q/At = σT^{4}

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} Wm^{2}/k^{4} emissivity of plate =1)

ANS 1000K

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

according to the Stefans law P = AeσT^{4}

so 1134 = 0.02*1*5.67*10^{-8}*T^{4 }(area of both the faces are considered)

so 1134*10^{10}/11.34 = T^{4}

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

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

thermal capacity = ms = Vds

so thermal capacity per unit volume = ds

now calculate answer

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

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

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

[x^{2}] α t

so 2^{2} - 1^{2} = k6

and 4^{2} - 3^{2 }= 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 .

**Asked By: SARIKA**

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**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 ?

**Asked By: SHUBHAM VED**

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**Joshi sir comment**

Rotational K. E. = 1/2 I ω^{2}

so 0.75 [1/2 I ω^{2}] = M S dT

and I = 2/3 M R^{2}

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

**Asked By: HIMANSHU MITTAL**

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**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(θ_{1}- θ_{2})/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

**Asked By: HIMANSHU MITTAL**

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**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 * T^{4}

so ratio Q_{a}:Q_{b}:Q_{c} = 1*(20)^{4} : 4*(15)^{4} : 9*(12)^{4}

on solving we will get that Q_{b }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

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

amount of energy given by bullet = 0.5 (1/2) m v^{2} / 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 v^{2}/(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

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

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/K4πr^{2}, on putting the value of K we get

dR = r^{2}dr/4πr^{2}k_{0}

or dR = dr/4πk_{0}

now integrate within the proper limits