Chapter 13, Problem 104A

To determine

Level of the water dropped or risen when the beaker and water are heated to $50°\text{C}$ .

Answer to Problem 104A

The water level will go down slightly.

Explanation of Solution

Given info:

Volume coefficient of expansion, $\beta$

${\beta }_{\text{water}}=210\times {10}^{-6}\text{per}°\text{C}$ and ${\beta }_{\text{beaker}}=27\times {10}^{-6}\text{per}°\text{C}$

Volume of water $\text{= 100}\text{mL}$

Volume of beaker $\text{= 800}\text{mL}$

Change in temperature, $\Delta T\text{= 35}°\text{C}$

Formula used:

Change in volume:

  $\text{Δ}V=\beta V\text{Δ}T$

Where, $\text{Δ}V$ is change in volume, $\beta$ is coefficient of volume expansion, $V$ is the volume, $\Delta T$ is the change in temperature.

Calculation:

Apply the formula $\text{Δ}V=\beta V\text{Δ}T$ and substitute all the given values:

Therefore, the water expands:

  $\Delta V\text{=}{\beta }_{\text{water}}V\Delta T$

  $\text{= }210\times {10}^{-6}\text{per}°\text{C}\times \text{100}\text{mL}\times \text{35}°\text{C}$

  $\text{=}\text{0}\text{.735}\text{mL}$

Therefore, the beaker expands: $\Delta V\text{=}{\beta }_{beaker}V\Delta T$

  $\text{=}27\times {10}^{-6}\text{per}°\text{C}\times \text{800}\text{mL}\times \text{35}°\text{C}$

  $\text{=}\text{0}\text{.756}\text{mL}$

Hence, the water level will drop.

Conclusion:

The water level will drop.