Two identical conducting rods are first connected independently to two vessels, one containing water at 100$^{\circ}$C and the other containing ice at 0$^{\circ}$C. In the second case, the rods are joined end to end and connected to the same vessels. Let q$_1$ and q$_2$ gram per second be the rate of melting of ice in the two cases respectively. The ratio $\frac{q_1}{q_2}$ is
- $\frac{1}{2}$
- $\frac{2}{1}$
- $\frac{4}{1}$
- $\frac{1}{4}$
The Correct Option is C
Solution and Explanation
or $ \, \, \, \frac{Temperature \, difference}{Thermal \, resistance} =L\bigg(\frac{dm}{dt}\bigg)$
or $\frac{dm}{dt} \propto \frac{1}{Thermal \, resistance} \Rightarrow q $
In the first case rods are in parallel and thermal resistance is
$\frac{R}{2}$ while in second case rods are in series and thermal
resistance is 2R.
$\frac{q_1}{q_2}=\frac{2R}{R / 2 }=\frac{4}{1}$
Hence, the correct option is (c)
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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