a researcher is cooling a metal. she cools the metal so that the temperature of the metal drops at a…

a researcher is cooling a metal. she cools the metal so that the temperature of the metal drops at a constant rate. after 19 minutes of cooling, the metal is 401 °c. after 42 minutes, the metal is 148 °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 cooling it?\n °c

a researcher is cooling a metal. she cools the metal so that the temperature of the metal drops at a constant rate. after 19 minutes of cooling, the metal is 401 °c. after 42 minutes, the metal is 148 °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 cooling it?\n °c

Answer

Explanation:

Step1: Calculate the rate of temperature change

The change in temperature is $\Delta T=401 - 148=253^{\circ}C$, and the change in time is $\Delta t = 42-19 = 23$ minutes. The rate of change of temperature $r=\frac{\Delta T}{\Delta t}$. $r=\frac{401 - 148}{42 - 19}=\frac{253}{23}=11^{\circ}C$ per minute. Since the temperature is dropping, as time increases, the temperature of the metal decreases, and the rate is $11^{\circ}C$ per minute.

Step2: Find the initial temperature

Let the initial temperature be $T_0$. We know that after $t = 19$ minutes, the temperature $T=401^{\circ}C$. Using the formula $T=T_0-rt$, we can solve for $T_0$. Rearranging the formula gives $T_0=T + rt$. Substitute $T = 401$, $r = 11$ and $t=19$ into the formula. $T_0=401+11\times19=401 + 209=610^{\circ}C$

Answer:

(a) As time increases, the temperature of the metal decreases. The temperature of the metal decreases at a rate of $11^{\circ}C$ per minute. (b) $610^{\circ}C$