Question

a researcher is heating a metal. he heats the metal so that the temperature of the metal rises at a constant rate. after 30 minutes of heating, the metal is 446 °c. after 50 minutes, the metal is 726 °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 (square) °c per minute.

• as time increases, the temperature of the metal increases.

the temperature of the metal increases at a rate of (square) °c per minute.
(b) what was the temperature of the metal when the researcher started heating it?
(square) °c

Explanation:

Part (a)

Step1: Determine temperature change

The temperature at 30 minutes is \(446^\circ\text{C}\) and at 50 minutes is \(726^\circ\text{C}\). The time difference is \(50 - 30 = 20\) minutes. The temperature change is \(726 - 446 = 280^\circ\text{C}\).

Step2: Calculate rate of increase

The rate of temperature increase is \(\frac{280}{20}=14^\circ\text{C}\) per minute. Since the temperature is rising, we choose the statement: "As time increases, the temperature of the metal increases. The temperature of the metal increases at a rate of \(14^\circ\text{C}\) per minute."

Part (b)

Step1: Use the rate to find initial temperature

Let the initial temperature (at time \(t = 0\)) be \(T_0\). At \(t = 30\) minutes, the temperature \(T = 446^\circ\text{C}\). The rate \(r = 14^\circ\text{C}\) per minute. Using the formula \(T = T_0+rt\), we can solve for \(T_0\).

Step2: Solve for \(T_0\)

Rearranging the formula: \(T_0=T - rt\). Substituting \(T = 446\), \(r = 14\), and \(t = 30\): \(T_0=446-(14\times30)=446 - 420 = 26^\circ\text{C}\).

Answer:

(a) As time increases, the temperature of the metal increases. The temperature of the metal increases at a rate of \(\boldsymbol{14}\) \(^\circ\text{C}\) per minute.
(b) \(\boldsymbol{26}\)