During an all-night cram session, a student heats up a 0.360 liter (0.360 x 10- 3 m3) glass (Pyrex) beaker of cold coffee. Initially, the temperature is 18.1 °C, and the beaker is filled to the brim. A short time later when the student returns, the temperature has risen to 94.6 °C. The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?
dV = V0 â (t1 - t0)
therefore
0.00036*0.000214(of water?)(94.6-18.1)
=
dV = V1 - V0 = change in volume
= answer
is this correct or am I way off
Your equation for the change in volume is partially correct, but there are a few mistakes in the calculation. Let me explain the correct approach step-by-step.
To calculate the change in volume, you can use the formula:
ΔV = V₀ * β * ΔT
Where:
ΔV is the change in volume
V₀ is the initial volume of the coffee in the beaker (0.360 L or 0.360 * 10^(-3) m³ in this case)
β is the coefficient of volume expansion (the same as water)
ΔT is the temperature change (final temperature - initial temperature)
Now, let's substitute the given values into the equation:
ΔV = 0.360 * 10^(-3) * β * (94.6 - 18.1)
The coefficient of volume expansion for water is approximately 0.000214 (1/°C), so the correct formula should be:
ΔV = 0.360 * 10^(-3) * 0.000214 * (94.6 - 18.1)
Let's calculate this:
ΔV = 0.360 * 10^(-3) * 0.000214 * 76.5
Now, multiply the values together:
ΔV ≈ 0.00699 * 10^(-3) m³
Finally, simplify the result:
ΔV ≈ 6.99 * 10^(-6) m³
Therefore, approximately 6.99 * 10^(-6) cubic meters (m³) of coffee have spilled out of the beaker.
Just a reminder, make sure to double-check the accuracy of the given coefficient of volume expansion for water (β) if it was provided separately.