Question

a researcher is cooling a metal. she cools the metal so that the temperature of the metal drops at a constant rate. after 17 minutes of cooling, the metal is 486°c. after 38 minutes, the metal is 234°c. (a) choose the statement that best describes how the time and the temperature of the metal are related. then fill in the blank. as time increases, the temperature of the metal decreases. the temperature of the metal decreases at a rate of ▢°c per minute. 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? try again your answer is incorrect. incorrect

Explanation:

Step1: Calculate the rate of temperature change

The metal cools from \(486^{\circ}C\) to \(234^{\circ}C\) over \(38\) minutes. The change in temperature is \(\Delta T=486 - 234=252^{\circ}C\).
The rate of change (slope) \(m=\frac{\Delta T}{\Delta t}\). Using the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), where \(y\) is temperature and \(x\) is time. Here, \(y_1 = 486\), \(y_2=234\), \(x_1 = 0\), \(x_2 = 38\). So \(m=\frac{234 - 486}{38-0}=\frac{- 252}{38}=- 6.6316\approx - 6.63^{\circ}C\) per minute. So as time increases, the temperature of the metal decreases at a rate of approximately \(6.63^{\circ}C\) per minute.

Step2: Use the point - slope form to find the initial temperature

The equation of a line is \(y - y_1=m(x - x_1)\). We know \(m=-6.63\), \(x = 38\), \(y = 234\). Using \(y=mx + b\) (where \(b\) is the initial temperature when \(x = 0\)). Substitute \(x = 38\), \(y = 234\) into \(y=-6.63x + b\).
\(234=-6.63\times38 + b\).
First calculate \(-6.63\times38=-251.94\). Then \(b=234 + 251.94 = 485.94\approx486^{\circ}C\)

Answer:

(a) As time increases, the temperature of the metal decreases at a rate of \(6.63^{\circ}C\) per minute. (b) \(486^{\circ}C\)