UPSEE Solved Model Paper 5

Here, we have to find out the temperature difference across the slab A.

We have steady state
Rate of heat flow through copper = Rate of heat flow through steel
We have
The coefficient of thermal conductivity of a substance
k =

Qd

A(θ1−θ2)t

where Q - heat conducted normally
d = length of the rod
A - area of the cross section
Q1 and Q2 - Temperatures at two ends
i.e., Q =

kA(θ1−θ2)t

d

Let θ be the temperature at the junction
Therefore, rate of heat flow
P =

Q

t

=

kA(θ1−θ2)

d