A metal block of heat capacity `90J//.^(@)C` placed in a room at `25^(@)C` is heated electrically. The heater is switched off when the temperature reaches `35^(@)C`. The temperature of the block rises at the rate of `2^(@)C//s` just after the heater is switched on and falls at the rate of `0.2^(@)C//s` just after the heater is switched off. Assume Newton's law of cooling to hold (a) Find the power of the heater. (b) Find the power radiated by the block just after the heater is switched off. (c ) Find the power radiated by the block when the temperature of the block is `30^(@)C`. (d) Assuming that the power radiated at `30^(@)C` respresents the average value in the heating process, find the time for which the heater was kept on.

Answer

A, B, C, D

Explanation

In steady state conditions no heat is absorbed. The reate of loss of heat by conduction is equal to that of the supplied.
`m = 1 Kg`.
Power of Heater `= 20 W`,
Room temperature `= 20^@C`
(a) `H = (dtheta)/(dt) = P = 20 watt`
(b) By Newton's law of cooling
`(dtheta)/(dt) = K (theta - theta^@)`
`implies 20 = K (50 - 20), K = 2/3`
Again, `(dtheta)/(dt) = K (theta - theta^@)`
`= 2/3 xx (30 - 20) = 20/3 W`
(c) `((dtheta)/(dt))_(2) = 0, ((dtheta)/(dt))_(30) = 20/3`
`((dtheta)/(dt))_(avg) = 10/3`
`T = 5 min = 300 s`
Heat liberated `= 10/3 xx 300 = 1000 J`
`Net Heat absorbed = Heat supplied`
`- Heat Radiated`
`= 6000 - 1000`
`= 5000 J`
Now, `mDeltathetaS = 5000`
`S = (5000)/(mDeltatheta) = (5000)/(1 xx 10)`
`= 500 J-Kg^(1)^@C^(-1)`