Dependence of temperature on time 1550. Heating element problem

05.03.2020

Let's look at the next 8 problems B12 from the Unified State Examination in mathematics. There are 2 topics here: the temperature of the device and the efficiency of the heat engine. For variety, some of the problems in which quadratic equations occur will be solved through the discriminant (see lesson “Solving quadratic equations”), and some - through Vieta’s formulas (see lesson “Vieta’s theorem”).

Task. The dependence of temperature (in degrees Kelvin) on time (in minutes) for the heating element of a certain device was obtained experimentally and over the temperature range under study is given by the expression T ( t) = T 0 + at + bt 2, where T 0 = 340 K, a = 28 K/min, b = −0.2 K/min. It is known that at heater temperatures above 1000 K the device can deteriorate, so it must be turned off. Determine (in minutes) the longest time after starting work the device should be turned off.

Everything revolves around temperature, which changes according to the law: T (t) = T 0 + at + bt 2. We need to find out at what point this temperature will cross the 1000 K mark. Since the temperature T 0, as well as the coefficients a and b, are known to us, we will compose and solve the equation:

1000 = 340 + 28t − 0.2t 2 ;
0.2t 2 −28t + 660 = 0 - moved all terms to the left;
t 2 − 140t + 3300 = 0 - multiply both sides by 5.

Discriminant: D = 140 2 − 4 1 3300 = 6400 = 64 100. Obviously, the root of the discriminant is 80. The roots of the quadratic equation are:
t 1 = (140 + 80) : 2 = 110;
t 2 = (140 − 80) : 2 = 30.

It turns out that we have two candidates for the answer: the numbers 110 and 30. We need to find longest time, and that's why many people choose the answer 110.

But let's remember what these numbers mean. So, at time t = 30 minutes, as well as at time t = 110 minutes, the temperature crosses the critical mark of 1000 K - the same one after which the device can deteriorate. Roughly speaking, the device will deteriorate after 30 minutes and after 110.

Conclusion: the device must be turned off after 30 minutes, since by 110 minutes it will have been damaged for a long time.

Task. The dependence of temperature (in degrees Kelvin) on time (in minutes) for the heating element of some device was obtained experimentally and over the temperature range under study is given by the expression T (t) = T 0 + at + bt 2, where T 0 = 520 K, a = 22 K/min, b = −0.2 K/min. It is known that at heater temperatures above 1000 K the device can deteriorate, so it must be turned off. Determine (in minutes) the longest time after the start of work the device should be turned off.

The task is completely similar to the previous one - only the coefficients are different. We know the maximum permissible temperature, so we will compose and solve the equation:

1000 = 520 + 22t − 0.2t 2 ;
0.2t 2 − 22t + 480 = 0 - collected everything on the left;
t 2 − 110t + 2400 = 0 - multiply both sides by 5.

The problem was reduced to the reduced quadratic equation. According to Vieta's theorem:
t 1 + t 2 = −(−110) = 110;
t 1 · t 2 = 2400.

Obviously, the roots are: 80 and 30, because 80 + 30 = 110; 80 · 30 = 2400. We find that the maximum temperature will be reached after 30 minutes and after 80. Therefore, the device must be turned off after 30 minutes.

Task. The dependence of temperature (in degrees Kelvin) on time (in minutes) for the heating element of some device was obtained experimentally and over the temperature range under study is given by the expression T (t) = T 0 + at + bt 2, where T 0 = 800 K, a = 52 K/min, b = −0.4 K/min. It is known that at heater temperatures above 2000 K the device can deteriorate, so it must be turned off. Determine (in minutes) the longest time after starting work the device should be turned off.

The problem is similar to the previous one, so let's look at a short solution. It is precisely this amount of calculations that will be a sufficient justification for the answer in this Unified State Examination in mathematics.

2000 = 800 + 52t − 0.4t 2 ;
0.4t 2 − 52t + 1200 = 0;
t 2 − 130t + 3000 = 0 - divided everything by a factor of 0.4.

We solve through the discriminant: D = 130 2 − 4 1 3000 = 4900. Root of the discriminant: 70. Find the roots of the equation:
t 1 = (130 + 70) : 2 = 100;
t 2 = (130 − 70) : 2 = 30.

Of the two numbers, choose the smallest - this is again the number 30.

Task. The dependence of temperature (in degrees Kelvin) on time (in minutes) for the heating element of some device was obtained experimentally and over the temperature range under study is given by the expression T (t) = T 0 + at + bt 2, where T 0 = 280 K, a = 26 K/min, b = −0.2 K/min. It is known that at heater temperatures above 1000 K the device can deteriorate, so it must be turned off. Determine (in minutes) the longest time after the start of work the device should be turned off.

Everything is the same - we compose and solve the equation:

1000 = 280 + 26t − 0.2t 2 ;
0.2t 2 − 26t + 720 = 0 - moved all terms to one side;
t 2 − 130t + 3600 = 0 - multiply each term by 5.

This is a reduced quadratic equation that can be solved well using Vieta's theorem:
t 1 + t 2 = −(−130) = 130 = 90 + 40;
t 1 · t 2 = 3600 = 90 · 40.

From the given formulas it is obvious that the roots are 90 and 40. As before, you will have to choose the smallest root - the number 40. Because the device will not “live” until 90 minutes.

Task. The dependence of temperature (in degrees Kelvin) on time (in minutes) for the heating element of a certain device was obtained experimentally and over the temperature range under study is given by the expression T (t) = T 0 + at + bt 2, where T 0 = 1100 K, a = 36 K/min, b = −0.2 K/min. It is known that at heater temperatures above 2000 K the device can deteriorate, so it must be turned off. Determine (in minutes) the longest time after starting work the device should be turned off.

Again the clone problem, which comes down to the equation:

2000 = 1100 + 36t − 0.2t 2 ;
0.2t 2 − 36t + 900 = 0;
t 2 − 180t + 4500 = 0.

Here we have the equation again. According to Vieta's theorem:
t 1 + t 2 = −(−180) = 180 = 150 + 30;
t 1 t 2 = 4500 = 150 30.

Now the roots are obvious - these are the numbers 150 and 30. The answer will be the smallest number, i.e. The device must be turned off after 30 minutes.

At what minimum heater temperature T 1 will the efficiency of this engine be at least 60% if the refrigerator temperature T 2 = 200? Give your answer in degrees Kelvin.

First, let's simplify the original formula. Multiplying both sides of the equality by the variable T 1, we get:
η T 1 = (T 1 − T 2) 100.

We specifically removed the percentage sign, since in the final equation there cannot be any percentages - there are only numbers. According to the conditions of the problem, we know the efficiency η = 60% and the temperature of the refrigerator T 1 = 200. Substitute these numbers into the formula - we get the equation:
60 · T 1 = (T 1 − 200) · 100.

Please note: units of measurement are again not written. No percentages, no Kelvins - just regular numbers. In principle, the same should be done in all problems B12. It’s just that we haven’t focused on this point until now, but we need to work with percentages carefully.

So, we solve the equation:
60 · T 1 = (T 1 − 200) · 100;
60T 1 = 100T 1 − 20 000 - opened the brackets;
60T 1 − 100T 1 = −20 000 - collected all terms with T 1 on the left;
−40T 1 = −20,000;
T 1 = 500 - divided everything by −40.

As you can see, the problem has been reduced to a simple linear equation that has one root. This is very good, because, unlike quadratic equations, here you don’t have to think about which root to write down as the answer.

Task. The efficiency of a certain engine is determined by the formula:

At what minimum heater temperature T 1 will the efficiency of this engine be at least 60% if the refrigerator temperature T 2 = 400? Give your answer in degrees Kelvin.

The task is completely similar to the previous one. Let's transform the original formula and then substitute the known variables into it:

η · T 1 = (T 1 − T 2) · 100 - converted the formula;
60 · T 1 = (T 1 − 400) · 100 - substituted numbers;
60T 1 − 100T 1 = −40 000 - group the terms containing the variable T 1;
−40T 1 = −40,000;
T 1 = 1000 - divided both sides by a factor of −40.

Task. The efficiency of a certain engine is determined by the formula:

At what minimum heater temperature T 1 will the efficiency of this engine be greater than 80% if the refrigerator temperature T 2 = 100? Give your answer in degrees Kelvin.

Another clone task. Let me give you just a short solution:

η · T 1 = (T 1 − T 2) · 100 - converted formula;
80 · T 1 = (T 1 − 100) · 100 - substituted numbers;
80T 1 − 100T 1 = −10,000;
−20T 1 = −10 000;
T 1 = 500 is the answer.

The dependence of temperature (in degrees Kelvin) on time for the heating element of a certain device was obtained experimentally. Over the interval under study, the temperature is calculated using the formula , where is time in minutes, , K/min 2 , K/min. It is known that if the heater temperature exceeds K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.

The solution of the problem

This lesson discusses solving the problem of determining the maximum time after which it is necessary to turn off the device so that it does not deteriorate. It should be noted that the solution to this problem can be used as preparation for the Unified State Exam in mathematics.

During the solution, the formula for the dependence of temperature on time for the heating element of a certain device is used: , where is the time in minutes from the moment of switching on. According to the conditions of the problem, the temperature above which the device can deteriorate is known. To solve the problem, an inequality of the form is compiled. When solving this inequality, the values known by the condition are substituted into it, resulting in a quadratic inequality, which is solved by the method of intervals. To do this, zeros are marked on the number line and a “snake” is drawn from right to left and from top to bottom. The intervals lying above the numerical axis, satisfying the greater than or equal sign, are the solution to this inequality. Since time cannot be a negative value, the interval of values up to zero is discarded. The solution to the problem is the right boundary of the first interval of inequality values.

Solution.

a And b:

4 minutes after switching on, the device will heat up to 1600 K, and with further heating it may deteriorate. Thus, the device must be turned off after 4 minutes.

Answer: 4.

Answer: 4

Solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of parameters a and b:

4 minutes after switching on, the device will heat up to 1800 K, and then it will heat up and may deteriorate. Thus, the device must be turned off after 4 minutes.

Answer: 4.

Answer: 4

The dependence of temperature (in degrees Kelvin) on time for the heating element of a certain device was obtained experimentally. Over the studied interval, the temperature is calculated using the formula T(t) = T 0 + bt + at 2, Where t- time in minutes, T 0 = 1380 K, A= −15 K/min 2, b= 165 K/min. It is known that if the heater temperature exceeds 1800 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.

Solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

4 minutes after switching on, the device will heat up to 1800 K, and with further heating it may deteriorate. Thus, the device must be turned off after 4 minutes.

Answer: 4.

Answer: 4

For the heating element of a certain device, the dependence of the temperature on the operating time was experimentally obtained: where is the time in minutes. It is known that at a temperature of the heating element above 1650 K the device can deteriorate, so it must be turned off. What is the longest time after starting work that the device should be turned off? Express your answer in minutes.

Solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

4 minutes after switching on, the device will heat up to 1650 K, and with further heating it may deteriorate. Thus, the device must be turned off after 4 minutes.

Answer: 4.

Answer: 4

For the heating element of a certain device, the dependence of the temperature on the operating time was experimentally obtained: where is the time in minutes. It is known that at a temperature of the heating element above 1600 K, the device can deteriorate, so it must be turned off. What is the longest time after starting work that the device should be turned off? Express your answer in minutes.

Solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

1 minute after switching on, the device will heat up to 1600 K, and with further heating it may deteriorate. Thus, the device must be turned off after 1 minute

Answer: 1.

Answer: 1

Solution.

prototype.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

Answer: 2.

Answer: 4

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

Answer: 2.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b: K/min. It is known that if the heater temperature exceeds 1900 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.

Solution.

This task has not yet been solved, we present the prototype solution.

The dependence of temperature (in degrees Kelvin) on time for the heating element of a certain device was obtained experimentally and, over the temperature range under study, is determined by the expression , where is time in minutes, K, K/min, K/min. It is known that if the heater temperature exceeds 1760 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.
, Where t— time in minutes, K, K/min, K/min. It is known that if the heater temperature exceeds 1600 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

K/min. It is known that if the heater temperature exceeds 1700 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.

Solution.

This task has not yet been solved, we present the prototype solution.

The dependence of temperature (in degrees Kelvin) on time for the heating element of a certain device was obtained experimentally and, over the temperature range under study, is determined by the expression , where is time in minutes, K, K/min, K/min. It is known that if the heater temperature exceeds 1760 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.
, Where t— time in minutes, K, K/min, K/min. It is known that if the heater temperature exceeds 1550 K, the device may deteriorate, so it must be turned off. Determine the longest time after starting work you need to turn off the device. Express your answer in minutes.

Solution.

This task has not yet been solved, we present the prototype solution.

Let us find at what point in time after the start of work the temperature will become equal to K. The problem is reduced to solving the equation for given values of the parameters a And b:

2 minutes after switching on, the device will heat up to 1760 K, and with further heating it may deteriorate. Thus, the device must be turned off after 2 minutes.

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