Question 612·Hard·Algebra

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During a chemical experiment, the temperature of a solution decreases at a constant rate. Two minutes after the reaction begins, the temperature is 80 °C, and fourteen minutes after the reaction begins, the temperature is 38 °C. Which linear equation gives the temperature $T(t)$ , in degrees Celsius, as a function of time $t$ , in minutes, since the reaction began?

A constant rate of temperature change signals a linear model: its slope is the temperature change divided by the elapsed time. Pair each reading with its time, use Desmos to find the slope, then use one reading to find the temperature at time zero. A decrease needs a negative slope, and the first reading is not automatically the starting temperature.

Hints

Hint 1
In a linear function, each reading gives a point $(t,T(t))$ . Keep each time paired with its temperature. What two points do the readings give you?
Hint 2
The slope measures the temperature change per minute. Subtract the earlier temperature from the later one, and subtract the times in that same order. What sign should the slope have?
Hint 3
The intercept $b$ in $T(t)=mt+b$ is the temperature at time zero. The $80^\circ\mathrm{C}$ reading happens later. How can you use its time and the slope to find $b$ ?

Step-by-step

Find the rate, then the starting temperature
Step 1
Pair each time with its temperature
Time $t$ is measured from when the reaction began, so $T(2)=80$ gives the point $(2,80)$ , and $T(14)=38$ gives $(14,38)$ . The later reading is lower, so the rate should be negative.
Step 2
Find the cooling rate
The slope is the temperature change divided by the time change. Take both changes in the same order: type $\frac{38-80}{14-2}$ in Desmos. It prints $-3.5$ , so the temperature falls $3.5^\circ\mathrm{C}$ per minute.
Step 3
Find the temperature at the start
The intercept $b$ in $T(t)=mt+b$ is the temperature at $t=0$ . The first reading is the starting value only if it occurs at time zero. Here $80^\circ\mathrm{C}$ occurs at $t=2$ . Put that reading and the slope into the model: $80=-3.5(2)+b$ . Type $80\sim-3.5(2)+b$ in Desmos. It reports $b=87$ under PARAMETERS.
Step 4
Write the temperature function
Put the rate $m=-3.5$ and starting temperature $b=87$ into $T(t)=mt+b$ : $T(t)=-3.5t+87$ . This gives the solution's temperature in degrees Celsius $t$ minutes after the reaction began. Choice C.

Question 612·Hard·Algebra

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this