The specific volume of water is given quite accurately by the empirical relation
ln(v) = −6.7081 + 1.01257 ln(T) +
280.663/T
Use this relation to derive an equation for the coefficient of thermal expansion � and determine
the temperature at which this quantity is equal to zero. What is the significance of
this temperature?
To derive the equation for the coefficient of thermal expansion (β), we can use the definition of β, which is given by:
β = (1/v) * (∂v/∂T)
Now, let's differentiate the empirical relation for ln(v) with respect to T:
d/dT [ln(v)] = d/dT [-6.7081 + 1.01257 ln(T) + 280.663/T]
Using the chain rule, we have:
(1/v) * dv/dT = 1.01257/T - 280.663/T^2
Rearranging the equation, we get:
dv/dT = v * (1.01257/T - 280.663/T^2)
Now, substituting this equation in the definition of β:
β = (1/v) * (∂v/∂T)
β = (1/v) * [v * (1.01257/T - 280.663/T^2)]
β = 1.01257/T - 280.663/T^2
Therefore, the equation for the coefficient of thermal expansion (β) is:
β = 1.01257/T - 280.663/T^2
To determine the temperature at which β is equal to zero, we set β equal to zero and solve for T:
0 = 1.01257/T - 280.663/T^2
Rearranging the equation, we have:
280.663/T^2 = 1.01257/T
Cross-multiplying and rearranging further, we get:
280.663 = 1.01257T
T = 280.663 / 1.01257
Calculating this, we find:
T ≈ 277.153 degrees Celsius
This temperature is significant because it represents the point at which the coefficient of thermal expansion for water becomes zero. At this temperature, water experiences no change in volume with a change in temperature. This temperature is known as the "temperature of maximum density" for water, where its density is at its highest value.