29.2k views
2 votes
2. A spherical pot contains 0.75 L of hot water at an initial temperature of 95°C. The pot has an emissivity of 0.60, and the surroundings are at a temperature of 20°C. Calculate the water's rate of heat loss by

radiation.

A 15 W
B 18 W
C 20 W
D 10 W

User Ktm
by
8.0k points

1 Answer

2 votes

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.

User Oxcug
by
7.7k points