Question

If the temperature of water at 30∘C is raised by 1∘C, the water will expand. At approximately what initial temperature T will water expand by twice as much when raised by 1∘C? Express your answer numerically in degrees Celsiu

Answer 1

To solve this problem it will be necessary to obtain a graph of the change in the volumetric coefficient of water expansion, β vs. T. With this graph we will make the relationship between the two volumetric states.

The change in the Volume is given as,

$\Delta V = \beta V_0 T$

Here,

$V_0$ = Initial Volume

$\beta$ = Coefficient of volume expansion

$\Delta T$= Change in temperature

Determine the change in volume when the temperature is raised by 1°C from the initial temperature 30°C

$\Delta V = \beta V_0 (1\°C)$

And the change in volume when the temperature is raised from the initial temperature T is

$\Delta V' = \beta' V_0 \Delta T$

Replacing $2\Delta V$ for  $\Delta V'$ and 1°C for $\Delta T$ we have that

$2\Delta V = \beta' V_0 (1\°C)$

Equation at both states we have that

$2 (\beta V_0 (1\°C)) = \beta' V_0 (1\°C)$

$\beta' = 2\beta$

From the graphic for 30°C the value of $\beta$ is,

$\beta' = 2(300*10^{-6}\°C)$

$\beta' = 600*10^{-6}\°C$

Using again the graph the coefficient of Volume expansion is 70°C.

Therefore the initial temperature is 70°C