a researcher is heating a metal. he heats the metal so that the temperature of the metal rises at a constant…

a researcher is heating a metal. he heats the metal so that the temperature of the metal rises at a constant rate. after 25 minutes of heating, the metal is 507 °c. after 35 minutes, the metal is 697 °c.\n(a) choose the statement that best describes how the time and the temperature of the metal are related. then fill in the blank.\n○as time increases, the temperature of the metal decreases.\nthe temperature of the metal decreases at a rate of °c per minute.\n○as time increases, the temperature of the metal increases.\nthe temperature of the metal increases at a rate of °c per minute.\n(b) what was the temperature of the metal when the researcher started heating it?\n °c

a researcher is heating a metal. he heats the metal so that the temperature of the metal rises at a constant rate. after 25 minutes of heating, the metal is 507 °c. after 35 minutes, the metal is 697 °c.\n(a) choose the statement that best describes how the time and the temperature of the metal are related. then fill in the blank.\n○as time increases, the temperature of the metal decreases.\nthe temperature of the metal decreases at a rate of °c per minute.\n○as time increases, the temperature of the metal increases.\nthe temperature of the metal increases at a rate of °c per minute.\n(b) what was the temperature of the metal when the researcher started heating it?\n °c

Answer

Answer:

(a) As time increases, the temperature of the metal increases. The temperature of the metal increases at a rate of 19 °C per minute. (b) 52 °C

Explanation:

Step1: Calculate the rate of temperature increase

The change in temperature is $\Delta T=697 - 507=190$ °C, and the change in time is $\Delta t = 35 - 25 = 10$ minutes. The rate of temperature increase $r=\frac{\Delta T}{\Delta t}=\frac{190}{10}=19$ °C per minute.

Step2: Find the initial - temperature

Let the initial temperature be $T_0$. We know that after $t = 25$ minutes, the temperature $T = 507$ °C. Using the formula $T=T_0+rt$, we substitute $T = 507$, $r = 19$, and $t = 25$. So $507=T_0+19\times25$. Then $T_0=507-19\times25=507 - 475=52$ °C.