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?

Question

1060 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-27T11:18:30

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

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

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,

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

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