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?

Hello

This is an example of charles's law equation.

This law established that when the temperature increases, the volume also increases in a proportional relationship.

Let's see the equation:

V1*T2=V2*T1

Where:

V1= initial volume

V2= final volume

T1= Initial temperature

T2=FInal temperature

So, let's suppose that we have an original volume of 2 L and, where the temperature raised by 1°C it expand by twice to 4 L:

Replacing the values of charles's equation as follow:

V1= 2L

V2 =4L

T1 = x (because we don't know the value yet... )

and, as we want to raise the temperature by 1°C:

T2= x +1°C:

Going back to the charles's equation and replacing:

2L*(x+1°C)=4L*x

Then, we have to solve the equation as follow:

2x+2=4x

All terms with x has to be on one side of the equation:

2=4x-2x

2=2x

x=1°C

So the answer is initial temperature= 1°C.

Let's probe this anwer, using charles's law equation with any volume, for example V1= 5L

5L*2°C=V2*1°C

Solving:

V2=5L*2°C/1°C

V2=10L

So for ANY volume value, to expand by twice as much when raised by 1°C the initial temperature has to be 1°C.

idk i’m struggling too

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.