A 20 W heater is used to heat a metal ball of mass 1 kg in a room...

At steady state, the rate of heat loss to the surroundings equals the power supplied by the heater.

$P = k(T_{\text{steady}} - T_0)$

$20 = k(50 - 20)$

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

During the $5$-minute interval, the temperature of the ball increases uniformly from 20^{}^{{}^{\circ}}\text{C} to 30^{}^{{}^{\circ}}\text{C}. Since the temperature change is uniform, the average temperature of the ball during this time is:

T_{\text{avg}} = \dfrac{20 + 30}{2} = 25^{}^{{}^{\circ}}\text{C}

The average rate of heat loss to the surroundings during this interval is:

$P_{\text{loss}} = k(T_{\text{avg}} - T_0) = \dfrac{2}{3}(25 - 20) = \dfrac{10}{3} \text{ W}$

The total time interval is $t = 5 \text{ min} = 300 \text{ s}$.

Total heat lost to the surroundings during this time is:

$H_{\text{loss}} = P_{\text{loss}} \times t = \dfrac{10}{3} \times 300 = 1000 \text{ J}$