A cylindrical rod of heat capacity `120 J//K` in a room temperature `27^(@)C` is heated internally by heater of power `250W`, The steady state temperature attained by the rod is `37^(@)C`. Fin the following:
(a) The initial rate of increse in temperature
(b) The steady state rate of emission of radiant heat.
If the heater is switched off, find
(c) The initial rate of decrease is temperature
(d) The rate of decrease in temperature of the cylinder when its temperature falls to `31^(@)C` and
(e) The maximum amount of heat lost by the cylinder

Explanation

Initially when the rod is in tehrmal equilibrium with the room at `27^(@)C` from prevost's theory of exchanges, the rate of total radiant heat emitted by the rod to the surrounding is same as the rte of total radiant energy absorbed by it from the surrounding. So, as soon as the heater is switched on, the entire heat provided by it is complletely absorbed by the rod.
`c(dT)/(dt)P==` Power of heater
`120(J)/(K)((dT)/(dt))=250(J)/(s)`
`(dT)/(dt)=2.08 K//s`
(a) Thus, the initial rate of increase in temperture of the rod is `2.08 K//s`.
(b) In the steady state, the absorption of the radiant energy by the rod ceases, which evidently implies that the heat energy liberated by the heater per second is emitted by the rod per second.
So, the steady state rate of emission of radiant heat from the rod=power of heater `=250 W`.
(c) Immediately after the heater is switched off, the rod continues emitting heat energy at the same rate, as during the steady state, by virtue of its temperature However, the heater stops supplying the heat to the rod completely. As a result, the temperature of the rod starts falling.
From `-c(dT)/(dt)=(dQ)/(dt)`, we have
`(-dT)/(dt)=(250W)/(120J//l)=2.08 K//s`
(d) Since temperature difference between the rod and the room is not too large during the process, so Newton's law of cooling becomes valid.
`(-dT)/(dt)=K(T-T_(0))`
Initially, `((-dT)/(dt))=K(37-27)^(2)C=10 K^(@)C`
Finally, at `T=31^(@)C`
`((-dT)/(dt))_(f)=K(31-27)^(@)C=4K^(@)C` (ii)
From Eqs. (i) and (ii). We get
`((dT)/(dt))_(f)=(4)/(10)(2.08K//s)=0.83K//s` [because `((-dT)/(dt))_(i)=2.08 K//s]`
(e) Since a body cannot be cooled below the temperature of the surroundings, so maximum fall in temperature of the rod can be `10^(@)C`. So, the maximum heat that the rod can lose, after the heater is swiched off, by the time its temperature falls to that of the room, will be `c(DeltaT)_("max")`
`=120(J)/(K)(10K)=1200J`