During a chemistry experiment, the temperature of a liquid increases at a constant rate. T… — SAT

During a chemistry experiment, the temperature of a liquid increases at a constant rate. The temperature of the liquid is $24^\circ\text{C}$ at $5$ minutes after the experiment starts, and $48^\circ\text{C}$ at $15$ minutes after the experiment starts. The temperature $T$ , in degrees Celsius, of the liquid $t$ minutes after the experiment starts can be modeled by the equation $T = mt + b$ , where $m$ and $b$ are constants. What is the value of $b$ ?

Answer: 12

Answer

The correct answer is 12. By translating the given information into two points on the line,

(5, 24)

and

(15, 48)

, we can find the slope

m = \frac{48 - 24}{15 - 5} = 2.4

. Substituting this slope and the point

(5, 24)

into

T = mt + b

yields

24 = 2.4(5) + b

, which simplifies to

24 = 12 + b

. Solving for

b

gives

b = 12

Step-by-Step Solution

Identify the data points representing the relationship between time and temperature.

The two data points are

(t_1, T_1) = (5, 24)

and

(t_2, T_2) = (15, 48)

These points represent coordinates

(t, T)

on the line representing the temperature over time.

Calculate the slope

m

of the linear equation.

m = \frac{48 - 24}{15 - 5} = \frac{24}{10} = 2.4

The slope represents the constant rate of temperature increase per minute.

Substitute the slope

m

and the coordinates of one point into the equation

T = mt + b

to solve for the y-intercept

b

24 = 2.4(5) + b \implies 24 = 12 + b \implies b = 12

This determines the value of the constant

b

, which corresponds to the initial temperature of the liquid.

Key Concept

Determining the y-intercept of a linear model in two variables given two points.

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