AP EAPCET 25-Aug-2021 Shift 1 Paper

Let KA and KB be the thermal conductivity of material A and B, respectively.
Let T1 be the temperature of A and T2 be the temperature of B.
At steady state, rate of flow of heat is given by,

ΔQ

Δt

=

ΔQA

Δt

=

ΔQB

Δt

⇒

K′A(T2−T1)

lA+lB

=

KAA(T−T1)

lA

=

KBA(T2−T)

lB

...(i)
where K is thermal conductivity of both material and A and l are area and length of the material respectively.
Using Eq. (i), we get

KAA

lA

(T−T1)=

KBA(T2−T)

lB

⇒ 2KB(T−T1)=KB(T2−T) (∵lA=lB and KA=2KB)
⇒ 2T−2T1=T2−T
⇒ 3T=T2+2T1
⇒T=

2T1+T2

3

Again using Eq. (i). we get

K′A(T2−T1)

lA+lA

=

KAA(T−T1)

lA

⇒

K′(T2−T1)

2JA

=

KA( 2T1+T2 3 −T1)

IA

⇒

K′(T2−T1)

2

=

KA(T2−T1)

3

⇒ K′=

2

3

KA=

2

3

×2K2
⇒ K′=

4

3

Ka
Using Eq. (i) to evaluate (T2−T),

K′A(T2−T1)

lA+lB

=

KBA(T2−T)

lB

⇒

K′A(T2−T1)

2lA

=

KBA(T2−T)

lA

⇒ (

4

3

KB)

A(T2−T1)

2

=KBA(T2−T)
⇒

2

3

(T2−T1)=T2−T
⇒ T2−T=

2

3

(T2−T1)=

2

3

×24°=16°C