# 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

$\setminus \Delta T = \frac{\setminus \Delta V}{\setminus \gamma V} = \frac{2 m l}{198 m l \setminus \times \left(4.4 \setminus \times {10}^{- 4} \text{per} \setminus {\quad}^{o} C\right)} = {22.95}^{o} C$

${T}_{f} = {T}_{i} + \setminus \Delta T = {30}^{o} C + {22.95}^{o} C = {52.95}^{o} C .$

#### 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:
$\setminus \gamma \setminus \equiv \frac{1}{V} \frac{\setminus \Delta V}{\setminus \Delta T}$ ...... (1)

$\setminus \Delta V$ : Change in volume for a temperature change of $\setminus \Delta T$,
$V \setminus q \quad$ : Initial volume,
$\setminus \gamma \setminus q \quad$ : Coefficient of volume expansion.

To fill the beaker the volume of water must expand from $198$ $m l$ to $200$ $m l$, which is a volume change of $\mathrm{dV} = 2$ $m l$.

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

$\setminus \Delta T = \frac{\setminus \Delta V}{\setminus \gamma V} = \frac{2 m l}{198 m l \setminus \times \left(4.4 \setminus \times {10}^{- 4} \text{per} \setminus {\quad}^{o} C\right)} = {22.95}^{o} C$

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

${T}_{f} = {T}_{i} + \setminus \Delta T = {30}^{o} C + {22.95}^{o} C = {52.95}^{o} C .$