Answer:
Step-by-step explanation:
Under steady state condition heat released to the room=heat dissipated out of the room Let θ be the temperature of heater Then
θ−20=α[20−(−20)] .(i)
and θ−10=α[10−(−40)] .(ii) Solving Eqs. (i) and (ii) we get
we get
θ=60∘C