Question

a researcher is cooling a metal. she cools the metal so that the temperature of the metal drops at a constant rate. after 11 minutes of cooling, the metal is 321 °c. after 21 minutes, the metal is 131 °c.
(a) choose the statement that best describes how the time and the temperature of the metal are related. then fill in the blank.
o as time increases, the temperature of the metal decreases.
the temperature of the metal decreases at a rate of °c per minute.
o as time increases, the temperature of the metal increases.
the temperature of the metal increases at a rate of °c per minute.
(b) what was the temperature of the metal when the researcher started cooling it?
°c

Explanation:

Step1: Calculate the rate of temperature - drop

The initial temperature is not given yet. We know that in 21 minutes, the temperature drops from an unknown initial temperature to 131°C. The change in temperature $\Delta T$ in 21 minutes is related to the rate of change. Since the temperature drops at a constant rate, the rate $r$ of temperature - drop is given by the formula $r=\frac{\text{Change in temperature}}{\text{Change in time}}$. The change in temperature is the difference between the initial and final temperatures. Let the initial temperature be $T_0$ and the final temperature $T = 131^{\circ}C$ after $t = 21$ minutes. The rate of temperature - drop $r=\frac{T_0 - 131}{21}$. But we can also use the fact that as time increases, the temperature decreases. The rate of temperature decrease is $r=\frac{321 - 131}{21}=\frac{190}{21}\approx9.05^{\circ}C$ per minute.
So, the statement "As time increases, the temperature of the metal decreases. The temperature of the metal decreases at a rate of $\frac{190}{21}\approx9.05^{\circ}C$ per minute" is correct.

Step2: Find the initial temperature

We use the linear - relationship formula $T=T_0-rt$. We know that $T = 131^{\circ}C$, $t = 21$ minutes and $r=\frac{190}{21}^{\circ}C$ per minute.
Substitute into the formula: $131=T_0-\frac{190}{21}\times21$.
Simplify the right - hand side: $131=T_0 - 190$.
Solve for $T_0$: $T_0=131 + 190=321^{\circ}C$.

Answer:

(a) As time increases, the temperature of the metal decreases. The temperature of the metal decreases at a rate of $\frac{190}{21}\approx9.05^{\circ}C$ per minute.
(b) $321^{\circ}C$