Temperature 61484

The air bubble at the bottom of the lake at a depth of h = 21 m has a radius of r1 = 1 cm at a temperature of t1 = 4 °C. The bubble rises slowly to the surface, and its volume increases. Calculate its radius when it reaches the lake's surface, with a temperature of t2 = 27 °C. Atmospheric pressure is b = 0.1 MPa, water density = 1030 kg/m³. Do not take surface tension into account.

Correct answer:

r₂ =  1.0272 cm

Step-by-step explanation:

$$\begin{gathered} h_1=21\text{m} \\ {T}_1=4°C\to K=4+273.16K=277.16K \\ {r}_1=1\text{cm}\to \text{m}=1:100\text{m}=0.01\text{m} \\ \rho =1030{\text{kg/m}}^3 \\ {V}_1=\frac{4}{3}\cdot \pi \cdot r_1^3=\frac{4}{3}\cdot 3.1416\cdot 0.01^3\doteq 4.1888\cdot 10^{-6}{\text{m}}^3 \\ {T}_2=27°C\to K=27+273.16K=300.16K \\ {p}_2=0.1\cdot 10^6=100000\text{Pa} \\ m=\rho \cdot V_1=1030\cdot 4.1888\cdot 10^{-6}\doteq 0.0043\text{kg} \\ {p}_1=p_2+m\cdot \rho \cdot h_1=100000+0.0043\cdot 1030\cdot 21\doteq 100093.3216\text{Pa} \\ \frac{p_1\cdot V_1}{T_1}=\frac{p_2\cdot V_2}{T_2} \\ {V}_2=\frac{p_1\cdot V_1\cdot T_2}{p_2\cdot T_1}=\frac{100093.3216\cdot 4.1888\cdot 10^{-6}\cdot 300.16}{100000\cdot 277.16}\doteq 4.5406\cdot 10^{-6}{\text{m}}^3 \\ {V}_2=\frac{4}{3}\cdot \pi \cdot r_2^3 \\ x=\sqrt[3]{\frac{3\cdot V_2}{4\pi }}=\sqrt[3]{\frac{3\cdot 4.5406\cdot 10^{-6}}{4\cdot 3.1416}}\doteq 0.0103\text{m} \\ {r}_2=x\to \text{cm}=x\cdot 100\text{cm}=0.0103\cdot 100\text{cm}=1.027\text{cm}=1.0272\text{cm} \end{gathered}$$

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