A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227^ ^...

The surface area of the cubical block is $A = 6a^2$.

Substituting $a = 0.05 \text{ m}$, we get $A = 6 \times (0.05)^2 = 0.015 \text{ m}^2$.

According to the Stefan-Boltzmann law, the net rate of heat loss is:

$\dfrac{dQ}{dt} = \sigma A (T^4 - T_0^4)$

The rate of decrease of temperature is given by:

$\dfrac{dT}{dt} = \dfrac{1}{mc} \dfrac{dQ}{dt} = \dfrac{\sigma A (T^4 - T_0^4)}{mc}$

Given values are:

$\sigma = 5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$

$T = 227^{{}^{{}^{\circ}}} \text{C} = 500 \text{ K}$

$T_0 = 27^{{}^{{}^{\circ}}} \text{C} = 300 \text{ K}$

$m = 1.0 \text{ kg}$

$c = 400 \text{ J kg}^{-1} \text{ K}^{-1}$

Substituting these values into the equation:

$\dfrac{dT}{dt} = \dfrac{5.67 \times 10^{-8} \times 0.015 \times (500^4 - 300^4)}{1.0 \times 400}$

$\dfrac{dT}{dt} = \dfrac{5.67 \times 10^{-8} \times 0.015 \times (625 \times 10^8 - 81 \times 10^8)}{400}$

$\dfrac{dT}{dt} = \dfrac{5.67 \times 0.015 \times 544}{400}$

$\dfrac{dT}{dt} = \dfrac{46.2672}{400} \approx 0.1156 ^{{}^{{}^{\circ}}}\text{C s}^{-1}$

Rounding to two significant figures, the rate of temperature decrease is $0.12 ^{{}^{{}^{\circ}}}\text{C s}^{-1}$.