A metal ball of mass 1 kg is placed in a room at 20^ ^ ^ C and...

Rate of heat loss to the surroundings is given by Newton's law of cooling:

$\dfrac{dQ_{\text{loss}}}{dt} = K(T - T_0)$

At steady state, the temperature is constant at T_s = 50^{}^{{}^{\circ}} \text{C}, so the heat supplied by the heater equals the heat lost to the surroundings:

$P = K(T_s - T_0)$

20 = K(50 - 20) \Rightarrow K = \dfrac{20}{30} = \dfrac{2}{3} \text{ W } ^{}^{{}^{\circ}} \textC^{-1}

During the heating from 20^{}^{{}^{\circ}} \text{C} to 30^{}^{{}^{\circ}} \text{C}, the average temperature of the ball is:

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

Since the temperature rises uniformly, the average rate of heat loss is evaluated at the average temperature:

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

The net average power used to heat the ball is:

$P_{\text{net}} = P - P_{\text{loss}} = 20 - \dfrac{10}{3} = \dfrac{50}{3} \text{ W}$

The heat required to raise the temperature of the ball by \Delta T = 10^{}^{{}^{\circ}} \text{C} is:

$Q = m c \Delta T = 1 \times c \times 10 = 10c$

This heat is supplied by the net power in time $t = 5 \text{ minutes} = 300 \text{ s}$:

$Q = P_{\text{net}} \times t$

$10c = \dfrac{50}{3} \times 300$

$10c = 5000 \Rightarrow c = 500 \text{ J kg}^{-1} \text{ K}^{-1}$