Inside a room at 20^ ^ ^ C , a metal ball of mass 1 kg is heated by...

At steady state, the rate of heat supplied by the heater is equal to the rate of heat lost to the surroundings.

According to Newton's law of cooling, the rate of heat loss is proportional to the temperature difference between the body and the surroundings:

$\dfrac{dQ}{dt} = K(T - T_0)$

Given that the heater supplies $20 \text{ W}$ and the steady-state temperature is 50^{}^{{}^{\circ}} \text{C} in a room at 20^{}^{{}^{\circ}} \text{C}:

$20 = K(50 - 20)$

20 = 30K \Rightarrow K = \dfrac{2}{3} \text{ W/}^{}^{{}^{\circ}} \text{C}

When the temperature of the ball is 30^{}^{{}^{\circ}} \text{C}, the rate of heat loss to the surroundings is:

$\dfrac{dQ'}{dt} = K(30 - 20) = 10K$

$\dfrac{dQ'}{dt} = 10 \times \dfrac{2}{3} = \dfrac{20}{3} \text{ W}$