What is the volume of water at 95 degrees Celcius if at 5 degrees Celcius the water is 5.56L?
To determine the volume of water at 95 degrees Celsius using the given information, you need to consider the concept of thermal expansion.
Thermal expansion refers to the tendency of matter to increase in volume as its temperature increases. This principle can be applied to calculate the final volume of the water.
One common equation used to calculate thermal expansion is the linear expansion equation:
ΔL = α * L * ΔT
Where:
- ΔL is the change in length,
- α is the coefficient of linear expansion,
- L is the original length, and
- ΔT is the change in temperature.
In this case, we want to calculate the change in volume. Since water is not a solid, we need to consider the coefficient of volume expansion, β, which is related to the coefficient of linear expansion, α, by the equation: β = 3α.
Given that the initial volume, V1, is 5.56L and the initial temperature, T1, is 5 degrees Celsius, and we want to find the final volume, V2, at a temperature of 95 degrees Celsius, we can use the formula:
V2 = V1 * (1 + β * ΔT)
Here's how to proceed with the calculations:
Step 1: Calculate the change in temperature
ΔT = T2 - T1
= 95°C - 5°C
= 90°C
Step 2: Calculate the coefficient of volume expansion, β
β = 3α
= 3 * α_water
The coefficient of volume expansion for water, α_water, is approximately 0.00021 (1/°C). Therefore:
β = 3 * 0.00021
= 0.00063 (1/°C)
Step 3: Calculate the final volume, V2
V2 = V1 * (1 + β * ΔT)
= 5.56L * (1 + 0.00063 * 90°C)
= 5.56L * (1 + 0.0567)
= 5.56L * 1.0567
≈ 5.86L
Therefore, the volume of water at 95 degrees Celsius would be approximately 5.86 liters.