Liquid water has a density of 1.00g/mL at 10.0C and .996g/mL at 30.0C. Calculate the change in volume that occurs when 500mL of water is heated from 10.0C to 30.0C.
volume = mass/density
Substitute and solve for volume for the two sets of temperature.
Well, let's start with some liquid comedy, shall we? Water is like a liquid chameleon - it likes to change its volume depending on the temperature. So, let's calculate the change that occurs when it goes from a chilly 10.0C to a steamy 30.0C!
First, we need to find the difference in density between the two temperatures. The density at 10.0C is a whopping 1.00g/mL, while at 30.0C it's a slimmer .996g/mL. So, the change in density is .004g/mL (1.00g/mL - .996g/mL).
Now comes the easy part. We just need to multiply the change in density by the original volume of 500mL. So, the change in volume is .004g/mL * 500mL = 2g.
Voila! The change in volume when heating 500mL of water from 10.0C to 30.0C is 2 grams. And that, my friend, is some liquid comedy gold!
To calculate the change in volume of water when heated from 10.0°C to 30.0°C, we will need to use the formula:
ΔV = Vf - Vi
where ΔV represents the change in volume, Vf is the final volume, and Vi is the initial volume.
Given:
Initial temperature (Ti) = 10.0°C
Final temperature (Tf) = 30.0°C
Initial density (ρi) = 1.00 g/mL
Final density (ρf) = 0.996 g/mL
Initial volume (Vi) = 500 mL
First, let's calculate the final volume (Vf) at 30.0°C using the final density formula:
Vf = M/ρf
where M is the mass.
To find the mass (M), we can use the formula:
M = ρi * Vi
Substituting the given values:
M = 1.00 g/mL * 500 mL = 500g
Now, substitute the values of M and ρf into the formula for final volume:
Vf = M/ρf = 500g / 0.996 g/mL = 502.01 mL
Using the formula for ΔV:
ΔV = Vf - Vi = 502.01 mL - 500 mL = 2.01 mL
Therefore, the change in volume that occurs when 500 mL of water is heated from 10.0°C to 30.0°C is 2.01 mL.