A geophysicist models the temperature, T, in degrees Celsius (^\circ\text{C}), of a rock l… — SAT

A geophysicist models the temperature, $T$ , in degrees Celsius ( $^\circ\text{C}$ ), of a rock layer during a deep-crust drilling project using a linear function of the depth, $d$ , in kilometers ( $\text{km}$ ), below the surface. According to the model, for every increase in depth of $0.8\text{ km}$ , the temperature of the rock increases by $22^\circ\text{C}$ . At a depth of $2.4\text{ km}$ , the temperature of the rock is $81^\circ\text{C}$ . According to the model, at what depth, in kilometers, will the temperature of the rock be $114^\circ\text{C}$ ?

Answer: 3.6 km

Answer

The temperature of the rock will reach

114^\circ\text{C}

at a depth of

3.6

kilometers.

The correct answer is

3.6

. The rate of change of temperature with depth is

\frac{22^\circ\text{C}}{0.8\text{ km}} = 27.5^\circ\text{C/km}

. The linear relationship between temperature

T

and depth

d

can be modeled by

T = 27.5d + T_0

, where

T_0

is the temperature at the surface. Substituting the known values

d = 2.4

and

T = 81

into the model gives

81 = 27.5(2.4) + T_0

, which simplifies to

81 = 66 + T_0

. Solving for

T_0

yields

T_0 = 15

. Thus, the model is

T = 27.5d + 15

. To find the depth when the temperature is

114^\circ\text{C}

, substitute

T = 114

into the model:

114 = 27.5d + 15

. Subtracting

15

from both sides gives

99 = 27.5d

, and dividing by

27.5

yields

d = 3.6

Step-by-Step Solution

Find the rate of change (slope) of the temperature with respect to depth.

The slope is

27.5^\circ\text{C/km}

The temperature increases by

22^\circ\text{C}

for every

0.8\text{ km}

of depth, so the rate of change is

\frac{22^\circ\text{C}}{0.8\text{ km}} = 27.5^\circ\text{C/km}

Set up a linear model and find the surface temperature (y-intercept).

The linear model is

T = 27.5d + 15

Using the slope-intercept form

T = md + T_0

and substituting the given values

d = 2.4

and

T = 81

gives

81 = 27.5(2.4) + T_0

, which simplifies to

81 = 66 + T_0

, so

T_0 = 15

Substitute the target temperature into the linear model and solve for the target depth.

d = 3.6

Substitute

T = 114

into the equation

T = 27.5d + 15

to get

114 = 27.5d + 15

. Subtracting

15

from both sides gives

99 = 27.5d

. Dividing both sides by

27.5

yields

d = 3.6

Key Concept

Interpreting slope and solving for values in a linear relationship context.

Alternative Method

Find the required temperature increase:

114^\circ\text{C} - 81^\circ\text{C} = 33^\circ\text{C}

. Set up a proportion using the rate of

22^\circ\text{C}

increase per

0.8\text{ km}

to find the change in depth

\Delta d

\frac{33^\circ\text{C}}{\Delta d} = \frac{22^\circ\text{C}}{0.8\text{ km}} \implies \Delta d = \frac{33 \times 0.8}{22} = 1.2\text{ km}

. Add this change in depth to the initial depth to find the final depth:

2.4 + 1.2 = 3.6\text{ km}

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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