Question

The rate of heat flow through a slab is H. If the slab thickness is halved, its cross-sectional area is doubled, and the temperature difference across it is doubled, then the rate of heat flow becomes (please show work)

Answer 1

Heat flow rate is given by:

H = k*A*dT/dx

k = thermal conductivity of slab

A = Cross-sectional Area of slab

dT = temperature difference across slab

dx = thickness of slab

Since 'k' for slab is constant, So

H2/H1 = (A2/A1)*(dT2/dT1)*(dx1/dx2)

Given that A2/A1 = 2

dT2/dT1 = 2

And dx2/dx1 = 1/2, So dx1/dx2 = 2

So,

H2/H1 = 2*2*2

H2 = 8*H1

So rate of heat flow across slab is increase by a factor of 8

Let me know if you've any query.

Answer 2

Solution: Rate of Heat Flow Under Modified Conditions

Given:

• Initial rate of heat flow ($H$):

  $$H=\frac{k\cdot A\cdot \mathrm{\Delta }T}{d}$$

  where:

• $k$ = thermal conductivity (constant),

• $A$ = cross-sectional area,

• $\mathrm{\Delta }T$ = temperature difference,

• $d$ = thickness of the slab.

• 
  

Modified Conditions:

1. Thickness ($d^{\mathrm{\prime }}$): Halved → $d^{\mathrm{\prime }}=\frac{d}{2}$.

2. Area ($A^{\mathrm{\prime }}$): Doubled → $A^{\mathrm{\prime }}=2A$.

3. Temperature difference ($\mathrm{\Delta }{T}^{\mathrm{\prime }}$): Doubled → $\mathrm{\Delta }{T}^{\mathrm{\prime }}=2\mathrm{\Delta }T$.

4. 
   

New Rate of Heat Flow ($H^{\mathrm{\prime }}$):

$$H^{\mathrm{\prime }}=\frac{k\cdot A^{\mathrm{\prime }}\cdot \mathrm{\Delta }{T}^{\mathrm{\prime }}}{d^{\mathrm{\prime }}}=\frac{k\cdot (2A)\cdot (2\mathrm{\Delta }T)}{d\mathrm{/}2}=\frac{4kA\mathrm{\Delta }T}{d\mathrm{/}2}=8\cdot \left(\frac{kA\mathrm{\Delta }T}{d}\right)=8H$$

Answer:

The new rate of heat flow becomes $\boxed{{\displaystyle 8H}}$.