The heat radiated per unit area in 1 hour by a furnace whose temperature is 3000 K is ($$\sigma = 5.7 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$$)

The furnace is assumed to behave like a perfect black body, so the Stefan-Boltzmann law can be used.

Radiation power (energy emitted per second) per unit area is given by
$$E = \sigma T^{4}$$

Given data: $$\sigma = 5.7 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$$ and $$T = 3000\ \text{K}$$.

First evaluate $$T^{4}$$:
$$T^{4} = (3000)^{4} = 3^{4}\times 10^{12} = 81 \times 10^{12} = 8.1 \times 10^{13}$$

Now compute the power per unit area:
$$E = \left(5.7 \times 10^{-8}\right)\left(8.1 \times 10^{13}\right)$$
$$E = 5.7 \times 8.1 \times 10^{\, -8 + 13}$$
$$E = 46.17 \times 10^{5}\ \text{W m}^{-2}$$
$$E = 4.617 \times 10^{6}\ \text{W m}^{-2}$$

Energy radiated per unit area in a given time is power multiplied by time. For 1 hour, the time is
$$t = 1\ \text{h} = 60 \times 60 = 3600\ \text{s}$$

Heat radiated per unit area:
$$Q = E \, t = \left(4.617 \times 10^{6}\right)\left(3600\right)$$
$$Q = 1.66212 \times 10^{10}\ \text{J m}^{-2}$$

Rounding to two significant figures,
$$Q \approx 1.7 \times 10^{10}\ \text{J m}^{-2}$$

Hence, the heat radiated per unit area in 1 hour is $$1.7 \times 10^{10}\ \text{J}$$.

Option A which is: $$1.7 \times 10^{10}$$ J