Answer:
B
Step-by-step explanation:
To calculate the rate of heat loss by radiation from the spherical pot of hot water, you can use the Stefan-Boltzmann law for radiation:
\[P = \varepsilon \cdot \sigma \cdot A \cdot (T_{\text{hot}}^4 - T_{\text{surroundings}}^4)\]
Where:
- \(P\) is the power or rate of heat loss.
- \(\varepsilon\) is the emissivity of the pot (0.60 in this case).
- \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\, \text{W/(m}^2\cdot \text{K}^4\text{)}\)).
- \(A\) is the surface area of the pot.
- \(T_{\text{hot}}\) is the temperature of the hot water in Kelvin (95°C + 273.15).
- \(T_{\text{surroundings}}\) is the temperature of the surroundings in Kelvin (20°C + 273.15).
The pot is spherical, and the surface area of a sphere is \(4\pi r^2\), where \(r\) is the radius. You'll need to calculate the radius using the volume of water in the pot.
Given:
- Volume of water (\(V\)) = 0.75 L = 0.75 dm³ = 0.75 dm³ = 0.75 kg (since 1 L of water is 1 kg).
- Density of water (\(\rho\)) = 1000 kg/m³ (approximately).
- Surface area of the pot (\(A\)) = \(4\pi r^2\).
First, calculate the radius (\(r\)) using the volume of water and assuming a spherical shape:
\[V = \frac{4}{3}\pi r^3\]
\[r = \left(\frac{3V}{4\pi}\right)^{1/3}\]
Now, plug in the values into the Stefan-Boltzmann equation:
\[\begin{align*}
P &= 0.60 \cdot 5.67 \times 10^{-8}\, \text{W/(m}^2\cdot \text{K}^4\text{)} \cdot 4\pi r^2 \cdot (T_{\text{hot}}^4 - T_{\text{surroundings}}^4) \\
P &= 0.60 \cdot 5.67 \times 10^{-8}\, \text{W/(m}^2\cdot \text{K}^4\text{)} \cdot 4\pi \left(\frac{3V}{4\pi}\right)^{2/3} \cdot \left((95 + 273.15)^4 - (20 + 273.15)^4\right)
\end{align*}\]
Calculate this to find the rate of heat loss (\(P\)):
\[P ≈ 17.9\, \text{W}\]
So, the rate of heat loss by radiation is approximately 18 W, which is option B.