Question

A 200 mL glass containing 198 mL of water at 30°C. The expansion coefficient of volume is 4,4 x `10^(-4)`°C. If the glass and water are heated, the water will expand and spill at what temperature?

Answer 1

`\DeltaT = (\DeltaV)/(\gammaV) = (2 ml)/(198ml\times(4.4\times10^{-4} "per"\quad ^oC)) = 22.95 ^oC`

`T_f = T_i + \DeltaT = 30^oC + 22.95^oC = 52.95^oC.`

Explanation:

Let us assume that the volume of glass does not change and it is only the water that undergoes thermal expansion.

The coefficient of volume expansion is defined as follows:
`\gamma \equiv 1/V(\DeltaV)/(\DeltaT)` ...... (1)

`\DeltaV` : Change in volume for a temperature change of `\DeltaT`,
`V\qquad` : Initial volume,
`\gamma\qquad` : Coefficient of volume expansion.

To fill the beaker the volume of water must expand from `198` `ml` to `200` `ml`, which is a volume change of `dV = 2` `ml`.

To calculate the increase in temperature that is required to accomplish this volume change, invert equation (1) and write it as,

`\DeltaT = (\DeltaV)/(\gammaV) = (2 ml)/(198ml\times(4.4\times10^{-4} "per"\quad ^oC)) = 22.95 ^oC`

So the temperature at which the water will start overflowing is,

`T_f = T_i + \DeltaT = 30^oC + 22.95^oC = 52.95^oC.`