Physics - Heat & Thermo Question with Solution | TestHub

The surface area of the sphere is $A=4\pi R^2$.$$A=4\pi (0.05\text{ m}{)}^2=0.01\pi {\text{ m}}^2$$

When the heater is switched off, the sphere cools due to net radiation loss to the enclosure walls. The rate of heat loss (power radiated) is given by Stefan-Boltzmann law for a black body in an enclosure:$$P_{loss}=\sigma A(T_{ss}^4-T_w^4)$$ Substituting the given values:$$P_{loss}=(5.67\times 10^{-8}{\text{ W/(m}}^2\cdot {\text{K}}^4))\times (0.01\pi {\text{ m}}^2)\times ((500\text{ K}{)}^4-(300\text{ K}{)}^4)$$ $$P_{loss}=(5.67\times 10^{-8})\times (0.01\pi )\times (625\times 10^8-81\times 10^8)$$ $$P_{loss}=(5.67\times 10^{-8})\times (0.01\pi )\times (544\times 10^8)\approx 96.91\text{ W}$$ The rate of cooling is related to the heat loss by $P_{loss}=-Mc\frac{dT}{dt}$. So, the initial rate of cooling $\left|\frac{dT}{dt}\right|$ is:$$\left|\frac{dT}{dt}\right|=\frac{P_{loss}}{Mc}=\frac{96.91\text{ W}}{(0.1\text{ kg})\times (400\text{ J/(kg}\cdot \text{K)})}$$ $$\left|\frac{dT}{dt}\right|=\frac{96.91}{40}\approx 2.4227\text{ K/s}$$ Rounding to two decimal places, the initial rate of cooling is $2.42$ K/s.