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Chapter 13: Problem 14

A cylindrical brass container with a base of \(75.0 \mathrm{cm}^{2}\) and height of \(20.0 \mathrm{cm}\) is filled to the brim with water when the system is at \(25.0^{\circ} \mathrm{C} .\) How much water overflows when the temperature of the water and the container is raised to \(95.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: Approximately \(16.17 \mathrm{cm}^3\) of water overflows.

Step by step solution

01

Find initial volume of the container

The initial volume of the container is given by $$V_{0} = \pi r^2 h$$. We are given the base area of the container \(A = 75.0 \mathrm{cm}^2\) instead of the radius. So, we can rewrite the formula as $$V_{0} = Ah$$. Substituting the given values: $$V_{0} = (75.0 \mathrm{cm}^2)(20.0 \mathrm{cm}) = 1500.0 \mathrm{cm}^3$$
02

Calculate the volume change for the container

Now, we will find the change in volume for the container due to the temperature change, using the volume expansion formula: $$\Delta V_{c} = V_{0} \beta_b \Delta T$$ Substituting the values: $$\Delta V_{c} = (1500.0 \mathrm{cm}^3)(60 \times 10^{-6} \mathrm{K}^{-1})(70.0 \mathrm{K}) = 6.3 \mathrm{cm}^3$$
03

Calculate the volume change for the water

Similarly, we'll find the change in volume for the water using the volume expansion formula: $$\Delta V_{w} = V_{0} \beta_w \Delta T$$ Substituting the values: $$\Delta V_{w} = (1500.0 \mathrm{cm}^3)(214 \times 10^{-6} \mathrm{K}^{-1})(70.0 \mathrm{K}) = 22.47 \mathrm{cm}^3$$
04

Find the volume of overflowed water

Finally, we'll find the volume of the overflowed water by subtracting the volume change of the container from the volume change of the water: $$V_{overflow} = \Delta V_{w} - \Delta V_{c}$$ $$V_{overflow} = 22.47 \mathrm{cm}^3 - 6.3 \mathrm{cm}^3 = 16.17 \mathrm{cm}^3$$ So, when the temperature of the water and the container is raised to \(95.0^{\circ} \mathrm{C}\), approximately \(16.17 \mathrm{cm}^3\) of water overflows.

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Show that, in two gases at the same temperature, the rms speeds are inversely proportional to the square root of the molecular masses: $$ \frac{\left(v_{\mathrm{rms}}\right)_{1}}{\left(v_{\mathrm{rms}}\right)_{2}}=\sqrt{\frac{m_{2}}{m_{1}}} $$
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