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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^

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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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