0.3

To find the emissivity of copper, we use the Stefan-Boltzmann law which states that the power radiated by a black body is given by:
$$ P = \varepsilon \sigma A T^4 $$
where:
- $P$ is the power radiated,
- $\varepsilon$ is the emissivity of the material,
- $\sigma$ is the Stefan-Boltzmann constant (approximately $5.67 \times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$),
- $A$ is the surface area of the sphere,
- $T$ is the absolute temperature in Kelvin.

**Step 1:** Determine the power needed to maintain the temperature of the sphere in two cases.
1. When the surface is not blackened (copper), power needed $P_c = 210 ext{ W}$.
2. When the surface is blackened (black body), power needed $P_{bb} = 700 ext{ W}$.

**Step 2:** Assume the surface area $A$ is the same in both cases. We denote the temperature of the sphere as $T = 500 K$.

**Step 3:** Using the Stefan-Boltzmann equation, we can set up two equations for both cases:
For copper:
$$ P_c = \varepsilon_c \sigma A T^4 $$
For black body:
$$ P_{bb} = \sigma A T^4 $$

**Step 4:** From the above equations, we can express $\varepsilon_c$:
$$ 210 = \varepsilon_c \cdot (5.67 \times 10^{-8}) imes A imes (500)^4 $$
$$ 700 = (5.67 \times 10^{-8}) imes A imes (500)^4 $$

**Step 5:** Now divide the two equations:
$$ \frac{210}{700} = \frac{\varepsilon_c}{1} $$
Thus,
$$ \varepsilon_c = \frac{210}{700} = 0.3 $$

Therefore, the emissivity of copper is 0.3.