Please enable scripting (or JavaScript) in your web browser, and then reload this page.
The half-life of Carbon 14 ($^14\C$) is 5730 years. This means that every 5730 years half of the remaining $^14\C$ in an organism disintegrates. For example, if there are 12 grams of $^14\C$ to start with, then after 5730 years there will be 6 grams remaining, and after the second 5730 years there will be half of the 6 grams, or 3 grams, remaining. Once something dies no more $^14\C$ is absorbed, so we can tell how long it has been dead by comparing how much $^14\C$ is currently in it to how much $^14\C$ was in it when it died.
Let $C$ be the amount of $^14\C$ left in an organism after $t$ years, if it had 1 gram of $^14\C$ when it died. We want to find an equation relating $C$ and $t$.
Every 5730 years, the amount of $^14\C$ is multiplied by $$1/2$$. This means that, every $t$ years, the amount of $^14\C$ is multiplied by $$(1/2)^{t∕5730}$$.
We actually want to use this equation to find out how long ago something died (like a fossil), so let’s look at it in a slightly different way. Suppose that you know that the $^14\C$ in a fossil is 0.1629 grams, and the amount that was in the organism when it died was 1.0 gram. Drag the vertical bar on the grid until $C$ has the value 0.1629. You see that $t = 15000$. This means that the organism died 15000 years ago.
A room with several large windows is normally kept at a temperature of 20 degrees Celsius (68 degrees Fahrenheit). The heat in the room is turned entirely off while the temperature outside is 5 degrees Celsius (41 degrees Fahrenheit). It’s then found that the temperature $y_1$ in the room $t$ hours later, in degrees Celsius, obeys the equation $y_1 = 15 ⋅ 2.33^{-t} + 5$. This equation has been graphed on the grid to the left.
When $t=0$ (that is, at the moment the heat is turned off), what is the temperature in the room in degrees Celsius?
In the room from tempQn1, the glass in the windows is replaced by double-glazed glass, which lets less heat through. It’s then found that, if the heat is turned off while the temperature outside is 5 degrees Celsius, the temperature $y_2$ in the room $t$ hours later obeys the equation $y_2 = 15 ⋅ 1.39^{-t} + 5$.
The two equations for temperature are both graphed on the grid to the left. The equation for the temperature with ordinary windows is graphed in red, and the equation for the temperature with double-glazed windows is graphed in blue.
Click to see what happens to the room over a longer time period (15 hours).
Now the two graphs on the left show what occurs if the heat in the room is turned off while the temperature outside is $-5$ degrees Celsius. Again, the red curve ($y_1$) shows how the room cools off if it has ordinary windows, and the blue curve ($y_2$) shows how it cools off if it has double-glazed windows.