A metal block having a heat capacity of 80 J ^ ^ ^ ^ -1 is located...

Let $P$ be the power of the heater and $k$ be the cooling constant.

Just after the heater is turned on at 20^{}^{{}^{\circ}}\text{C}, the temperature difference with the surroundings is zero, so the heat loss is zero. The rate of heat supplied is entirely used to increase the temperature:

$P = C \left(\dfrac{dT}{dt}\right)_{\text{initial}} = 80 \times 2 = 160 \text{ W}$

Just after the heater is turned off at 30^{}^{{}^{\circ}}\text{C}, the rate of heat loss is:

$P_{\text{loss}} = C \left(-\dfrac{dT}{dt}\right)_{\text{cooling}} = 80 \times 0.2 = 16 \text{ W}$

According to Newton's law of cooling, the rate of heat loss is proportional to the temperature difference, $P_{\text{loss}} = k(T - T_0)$. At 30^{}^{{}^{\circ}}\text{C}:

16 = k(30 - 20) \Rightarrow 10k = 16 \Rightarrow k = 1.6 \text{ W }^{}^{{}^{\circ}}\textC^{-1}

The average power radiated during the heating process is given as the power radiated at 25^{}^{{}^{\circ}}\text{C}:

$P_{\text{loss, avg}} = k(25 - 20) = 1.6 \times 5 = 8 \text{ W}$

The average net power available to heat the block is:

$P_{\text{net}} = P - P_{\text{loss, avg}} = 160 - 8 = 152 \text{ W}$

The total heat required to raise the temperature of the block from 20^{}^{{}^{\circ}}\text{C} to 30^{}^{{}^{\circ}}\text{C} is:

$\Delta Q = C \Delta T = 80 \times (30 - 20) = 800 \text{ J}$

The duration for which the heater was kept on is:

$t = \dfrac{\Delta Q}{P_{\text{net}}} = \dfrac{800}{152} = \dfrac{100}{19} \approx 5.26 \text{ s}$

Answer: $5.26 \text{ s}$