A laboratory technician is cooling a liquid sample. The initial temperature of the sample … — SAT

A laboratory technician is cooling a liquid sample. The initial temperature of the sample is $80^\circ\text{C}$ . The technician uses a cooling program that decreases the temperature at a constant rate of $1.5^\circ\text{C}$ per minute. After $t$ minutes, the technician increases the cooling rate by $0.75^\circ\text{C}$ per minute and runs the cooling program for another $12$ minutes. If the final temperature of the sample is $44^\circ\text{C}$ , what is the value of $t$ ?

A
8
6Cevap
C
18
D
42

Cevap

The correct answer is

6

. The initial temperature of the sample is

80^\circ\text{C}

. For the first

t

minutes, the temperature decreases by

1.5t^\circ\text{C}

. For the next

12

minutes, the rate is

1.5 + 0.75 = 2.25^\circ\text{C}

per minute, resulting in a temperature decrease of

2.25 \times 12 = 27^\circ\text{C}

. The final temperature equation is

80 - 1.5t - 27 = 44

, which simplifies to

53 - 1.5t = 44

. Solving for

t

gives

1.5t = 9

, or

t = 6

Adım Adım Çözüm

Determine the cooling rate for both phases of the program.

The initial rate is

1.5^\circ\text{C}

per minute. The increased rate is

1.5 + 0.75 = 2.25^\circ\text{C}

per minute.

The rate in the second phase is increased by

0.75^\circ\text{C}

per minute from the initial rate.

Set up the linear equation representing the temperature change over time.

80 - 1.5t - 2.25(12) = 44

The final temperature is the initial temperature minus the temperature drops during each phase.

Solve the equation to isolate the variable

t

80 - 1.5t - 27 = 44 \implies 53 - 1.5t = 44 \implies -1.5t = -9 \implies t = 6

Perform basic algebraic operations to find the value of

t

Anahtar Kavram

Setting up and solving a linear equation in one variable from a real-world scenario.

Soru

Cevap

Adım Adım Çözüm

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