Question:

Two rods of same material have same length and area. The heat $ \Delta Q $ flows through them for $12\, min$ when they are joint side by side. If now both the rods are joined in parallel, then the same amount of heat $ \Delta Q $ will flow in:

Updated On: Jun 20, 2022
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The Correct Option is B

Solution and Explanation

When two rods are joined, then the rate of flow of heat is given by
$ Q=KA\frac{({{\theta }_{1}}-{{\theta }_{2}})}{l}t $
where K is coefficient of thermal conductivity, A is area and I is length when rods are joined in series.
$ \Delta {{Q}_{1}}=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{2}}}{{{K}_{2}}}} $
Given, $ {{l}_{1}}={{l}_{2}}=l,{{K}_{1}}={{K}_{2}}=K, $
we have $ \Delta {{Q}_{1}}=\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{\frac{l}{{{K}_{1}}}+\frac{l}{{{K}_{2}}}} $
$ =\frac{A({{T}_{1}}-{{T}_{2}}){{t}_{1}}}{l}\frac{K}{2} $
when rods are joined in parallel
$ \Delta {{Q}_{2}}=({{K}_{1}}A+{{K}_{2}}A)\frac{({{T}_{1}}-{{T}_{2}}){{t}_{2}}}{l} $
$ =2\frac{KA({{T}_{1}}-{{T}_{2}}){{t}_{2}}}{l} $
Given, $ \Delta {{Q}_{1}}=\Delta {{Q}_{2}} $
$ \therefore $ $ {{t}_{2}}=\frac{{{t}_{1}}}{4}=\frac{12}{4}=3\,\min $
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Concepts Used:

Newton’s Law of Cooling

Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings. 

Derivation of Newton’s Law of Cooling

Let a body of mass m, with specific heat capacity s, is at temperature T2 and T1 is the temperature of the surroundings. 

If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,

dQ = ms dT2

The rate of loss of heat is given by,

dQ/dt = ms (dT2/dt)                                                                                                                                                                              ……..(2)

Compare the equations (1) and (2) as,

– ms (dT2/dt) = k (T2 – T1)

Rearrange the above equation as:

dT2/(T2–T1) = – (k / ms) dt

dT2 /(T2 – T1) = – Kdt 

where K = k/m s

Integrating the above expression as,

loge (T2 – T1) = – K t + c

or 

T2 = T1 + C’ e–Kt

where C’ = ec