Question

a researcher is heating a metal. he heats the metal so that the temperature of the metal rises at a constant rate. after 24 minutes of heating, the metal is 415°c. after 38 minutes, the metal is 639°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 heating it?
□°c

Explanation:

Step1: Find the rate of temperature increase

The temperature of the metal rises at a constant rate. After \(24\) minutes, the metal is \(415^{\circ}C\). After \(38\) minutes, the metal is \(639^{\circ}C\). The change in temperature is \(\Delta T=639 - 415=224^{\circ}C\), and the change in time is \(\Delta t = 38-24 = 14\) minutes. The rate \(r=\frac{\Delta T}{\Delta t}=\frac{639 - 415}{38 - 24}=\frac{224}{14}=16^{\circ}C\) per minute. So, as time increases, the temperature of the metal increases at a rate of \(16^{\circ}C\) per minute.

Step2: Find the initial temperature

We use the point - slope form of a linear equation \(T=T_0+rt\), where \(T\) is the temperature at time \(t\), \(T_0\) is the initial temperature, and \(r\) is the rate. Using \(t = 24\) minutes and \(T=415^{\circ}C\), \(r = 16^{\circ}C\) per minute. Then \(415=T_0+16\times24\). Solve for \(T_0\): \(T_0=415-16\times24=415 - 384=31^{\circ}C\)

Answer:

(a) As time increases, the temperature of the metal increases at a rate of \(16^{\circ}C\) per minute.
(b) \(31\)