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(D) Mass of the ball, $m = 1 \ kg$.Power of the heater, $P = 40 \ W$.Room temperature, $T_0 = 30^{\circ} C$.Steady state temperature of the ball, $T =

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$10$

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(D) Mass of the ball, $m = 1 \ kg$.Power of the heater, $P = 40 \ W$.Room temperature, $T_0 = 30^{\circ} C$.Steady state temperature of the ball, $T = 70^{\circ} C$.At steady state, the rate of heat supplied by the heater equals the rate of heat loss to the surroundings.According to Newton's law of cooling, the rate of heat loss is given by $\frac{dQ}{dt} = k(T - T_0)$.At steady state, $\frac{dQ}{dt} = P = 40 \ W$.Substituting the values: $40 = k(70 - 30) \Rightarrow 40 = k(40) \Rightarrow k = 1 \ W/^{\circ} C$.Now, we need to find the rate of heat loss when the ball is at $T_1 = 40^{\circ} C$.Using the same law: $\frac{dQ_1}{dt} = k(T_1 - T_0)$.Substituting the values: $\frac{dQ_1}{dt} = 1(40 - 30) = 10 \ W$.

$A$ metal ball of mass $1 \ kg$ is heated using a $40 \ W$ heater in a room at $30^{\circ} C$. The temperature of the ball becomes steady at $70^{\circ} C$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} C$ is (in $W$)

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