A heated solution is left to cool and its temperature decreases at a constant rate of degrees Celsius per minute. After minutes, the temperature of the solution is . What was the temperature, in degrees Celsius, of the solution when it was first left to cool?

Solution available with step-by-step explanation

SAT Linear Equations in One Variable Question 29 | Free SAT Prep | Aniko

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Question 29·Medium·Linear Equations in One Variable

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A heated solution is left to cool and its temperature decreases at a constant rate of $2.5$ degrees Celsius per minute. After $18$ minutes, the temperature of the solution is $37\,^{\circ}\text{C}$ . What was the temperature, in degrees Celsius, of the solution when it was first left to cool?

Your answer

(Express the answer as an integer)

For linear cooling or heating problems, first identify what is changing at a constant rate (here, temperature per minute) and what value is being asked for (usually the starting or ending amount). Compute the total change by multiplying the rate by the time, then decide whether to add or subtract that change based on whether you are moving backward or forward in time. Writing a simple equation like $T - (\text{rate} \times \text{time}) = \text{final temperature}$ and solving for $T$ keeps the work organized and reduces errors.

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Identify the unknown

What quantity are you trying to find: the temperature after 18 minutes, or the temperature at the moment it was first left to cool? Let that starting temperature be a variable, such as $T$ .

Find the total change in temperature

If the temperature decreases by $2.5$ degrees every minute, how many degrees total does it decrease in $18$ minutes? Multiply the rate by the time.

Decide whether to add or subtract

You know the temperature after it has cooled is $37$ °C. If you know how much it cooled, should the starting temperature be greater than or less than $37$ °C? Use that idea to decide whether to add or subtract the total decrease.

Desmos Guide

Use Desmos to compute the starting temperature

In Desmos, type the expression 37 + 2.5*18 and press Enter. The value Desmos outputs is the original temperature of the solution before it started cooling.

Step-by-step Explanation

Represent the situation

Let the initial temperature (when the solution was first left to cool) be $T$ degrees Celsius.

Because the temperature decreases (goes down) at $2.5$ degrees per minute for $18$ minutes, the total amount the temperature drops is

2.5 \times 18.

After this drop, the temperature is $37$ °C, so we can write the equation

T - (2.5 \times 18) = 37.

Find the total temperature decrease

Now calculate how much the temperature decreased in $18$ minutes:

$2.5 \times 10 = 25$
$2.5 \times 8 = 20$

2.5 \times 18 = 25 + 20 = 45.

Substitute this into the equation:

T - 45 = 37.

Solve for the initial temperature

To undo subtracting $45$ , add $45$ to both sides of the equation:

T - 45 + 45 = 37 + 45 \\[0.7em] T = 82.

So, the temperature of the solution when it was first left to cool was $82$ degrees Celsius.

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Step-by-step Explanation

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Step-by-step Explanation