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?

1 Answer
Nov 27, 2017

\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.