The room heater can provide only `16^(@)C` in the room when the temperature outside is `-20^(@)C`. It is not warm and comfortable, that is why th electric stove with power of `1 kW` is also plugged in. Together these two devices maintain the room temperature of `22^(@)C`. Determine the thermal power of the heater.

Explanation

Rate of heat loss with only room heater
`P_(h)=(DeltaQ)/(Deltat)=C(16+20)` [where `C=`constant]
while both heater and stove it is
`P_(h)+P_(s)=((DeltaQ)/(Deltat))=C(22+20)`
`:. (P_(h))/(P_(h)+P_(s))=(36)/(42)implies 7P_(h)=6P_(h)+6P_(s)`
`implies P_(h)=6P_(s)=6kW`