During a chemistry experiment, the temperature of a liquid sample is decreased at a consta… — SAT

During a chemistry experiment, the temperature of a liquid sample is decreased at a constant rate. At the start of the experiment, the temperature of the liquid is $80^\circ\text{C}$ . After $4$ minutes, the temperature of the liquid is $68^\circ\text{C}$ . If $T(t)$ represents the temperature of the liquid, in degrees Celsius, $t$ minutes after the experiment starts, which of the following equations defines $T$ ?

$T(t) = -3t + 80$ Answer
B
$T(t) = 3t + 80$
C
$T(t) = -\frac{1}{3}t + 80$
D
$T(t) = 80t - 3$

Answer

The equation

T(t) = -3t + 80

correctly defines the temperature function.

The correct equation shows a starting value of

80

and a rate of change of

-3

, which corresponds to the initial temperature of

80^\circ\text{C}

and a decrease of

3^\circ\text{C}

per minute.

Step-by-Step Solution

Identify the

y

-intercept (initial value) of the linear function.

The initial temperature at

t = 0

80^\circ\text{C}

, so the

y

-intercept

b

80

The initial value of a linear relationship in context corresponds to the

y

-intercept.

Calculate the slope (constant rate of change) using the coordinate points

(0, 80)

and

(4, 68)

The slope

m

\frac{68 - 80}{4 - 0} = \frac{-12}{4} = -3

The slope is defined as the change in the dependent variable (temperature) divided by the change in the independent variable (time).

Substitute the slope

m

and

y

-intercept

b

into the slope-intercept form

T(t) = mt + b

T(t) = -3t + 80

Substituting the specific slope and intercept values into the general form defines the particular function.

Key Concept

Determining a linear equation from a real-world scenario by identifying the initial value and constant rate of change.

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