00:01
For this question, both the steel tank and the ethanol undergo volumetric expansion.
00:07
So we'll need to use the equation, delta v, which the volume expansion, is equal to v0, the initial volume, multiplied by the coefficient of volumetric expansion, beta, times the change in temperature, delta t.
00:24
So first up, we do the volume change for the tank.
00:28
So we'll call it delta vs for the volume expansion of the temperature.
00:34
The steel and this is equal to v0 times beta for steel times delta t and so this is equal to the initial volume of the tank which is 2 .8 cubic meters times the volumetric expansion coefficient for steel from our table is 3 .6 times 10 to the minus 5 per celsius degree and the change in temperature is minus 40 degrees celsius.
01:21
So computing this we get that the volumetric change of the steel tank is minus 1 .41 times 10 to the minus 3 cubic meters which we can write as minus 1 .41 liters.
01:45
So the tank loses 1 .41 liters of capacity.
01:51
Now the volume change for ethanol, we'll call it delta ve, is equal to the initial volume of the ethanol multiplied by the volume coefficient, the expansion coefficient, beta for ethanol times the change in temperature delta t.
02:13
So initially the ethanol occupies the same volume, which is 2 .8 cubic meters.
02:23
The volumetric expansion coefficient for ethanol, more tables, is 75 times 10 to the minus 5 a celsius degree...