Newton's law of cooling

6. A hot piece of metal is dropped into a cool liquid. As the metal cools, its temperature \(T\) degrees Celsius, \(t\) minutes after it enters the liquid, is modelled by $$T = 300 \mathrm { e } ^ { - 0.04 t } + 20 , \quad t \geqslant 0$$

1. Find the temperature of the piece of metal as it enters the liquid.
2. Find the value of \(t\) for which \(T = 180\), giving your answer to 3 significant figures. (Solutions based entirely on graphical or numerical methods are not acceptable.)
3. Show, by differentiation, that the rate, in degrees Celsius per minute, at which the temperature of the metal is changing, is given by the expression $$\frac { 20 - T } { 25 }$$

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8. A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at \(3 ^ { \circ } \mathrm { C }\) and \(t\) minutes after the bottle is placed in the refrigerator the temperature of the water in the bottle is \(\theta ^ { \circ } \mathrm { C }\). The rate of change of the temperature of the water in the bottle is modelled by the differential equation, $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \frac { ( 3 - \theta ) } { 125 }$$

1. By solving the differential equation, show that, $$\theta = A \mathrm { e } ^ { - 0.008 t } + 3$$ where \(A\) is a constant. Given that the temperature of the water in the bottle when it was put in the refrigerator was \(16 ^ { \circ } \mathrm { C }\),
2. find the time taken for the temperature of the water in the bottle to fall to \(10 ^ { \circ } \mathrm { C }\), giving your answer to the nearest minute.

6. Water is being heated in a kettle. At time \(t\) seconds, the temperature of the water is \(\theta ^ { \circ } \mathrm { C }\). The rate of increase of the temperature of the water at any time \(t\) is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \lambda ( 120 - \theta ) , \quad \theta \leqslant 100$$ where \(\lambda\) is a positive constant. Given that \(\theta = 20\) when \(t = 0\),

1. solve this differential equation to show that $$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t }$$ When the temperature of the water reaches \(100 ^ { \circ } \mathrm { C }\), the kettle switches off.
2. Given that \(\lambda = 0.01\), find the time, to the nearest second, when the kettle switches off.

8. Water is being heated in a kettle. At time \(t\) seconds, the temperature of the water is \(\theta ^ { \circ } \mathrm { C }\). The rate of increase of the temperature of the water at time \(t\) is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \lambda ( 120 - \theta ) \quad \theta \leqslant 100$$ where \(\lambda\) is a positive constant.
Given that \(\theta = 20\) when \(t = 0\)

1. solve this differential equation to show that $$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t }$$ When the temperature of the water reaches \(100 ^ { \circ } \mathrm { C }\), the kettle switches off.
2. Given that \(\lambda = 0.01\), find the time, to the nearest second, when the kettle switches off.
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1. A bath is filled with hot water which is allowed to cool. The temperature of the water is \(\theta ^ { \circ } \mathrm { C }\) after cooling for \(t\) minutes and the temperature of the room is assumed to remain constant at \(20 ^ { \circ } \mathrm { C }\).

Given that the rate at which the temperature of the water decreases is proportional to the difference in temperature between the water and the room,

1. write down a differential equation connecting \(\theta\) and \(t\). Given also that the temperature of the water is initially \(37 ^ { \circ } \mathrm { C }\) and that it is \(36 ^ { \circ } \mathrm { C }\) after cooling for four minutes,
2. find, to 3 significant figures, the temperature of the water after ten minutes. Advice suggests that the temperature of the water should be allowed to cool to \(33 ^ { \circ } \mathrm { C }\) before a child gets in.
3. Find, to the nearest second, how long a child should wait before getting into the bath.

9 A liquid is being heated in an oven maintained at a constant temperature of \(160 ^ { \circ } \mathrm { C }\). It may be assumed that the rate of increase of the temperature of the liquid at any particular time \(t\) minutes is proportional to \(160 - \theta\), where \(\theta ^ { \circ } \mathrm { C }\) is the temperature of the liquid at that time.

1. Write down a differential equation connecting \(\theta\) and \(t\). When the liquid was placed in the oven, its temperature was \(20 ^ { \circ } \mathrm { C }\) and 5 minutes later its temperature had risen to \(65 ^ { \circ } \mathrm { C }\).
2. Find the temperature of the liquid, correct to the nearest degree, after another 5 minutes. 4

9 The temperature of a freezer is \(- 20 ^ { \circ } \mathrm { C }\). A container of a liquid is placed in the freezer. The rate at which the temperature, \(\theta ^ { \circ } \mathrm { C }\), of a liquid decreases is proportional to the difference in temperature between the liquid and its surroundings. The situation is modelled by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta + 20 ) ,$$ where time \(t\) is in minutes and \(k\) is a positive constant.

1. Express \(\theta\) in terms of \(t , k\) and an arbitrary constant. Initially the temperature of the liquid in the container is \(40 ^ { \circ } \mathrm { C }\) and, at this instant, the liquid is cooling at a rate of \(3 ^ { \circ } \mathrm { C }\) per minute. The liquid freezes at \(0 ^ { \circ } \mathrm { C }\).
2. Find the value of \(k\) and find also the time it takes (to the nearest minute) for the liquid to freeze. The procedure is repeated on another occasion with a different liquid. The initial temperature of this liquid is \(90 ^ { \circ } \mathrm { C }\). After 19 minutes its temperature is \(0 ^ { \circ } \mathrm { C }\).
3. Without any further calculation, explain what you can deduce about the value of \(k\) in this case.

9 A tank contains water which is heated by an electric water heater working under the action of a thermostat. The temperature of the water, \(\theta ^ { \circ } \mathrm { C }\), may be modelled as follows. When the water heater is first switched on, \(\theta = 40\). The heater causes the temperature to increase at a rate \(k _ { 1 } { } ^ { \circ } \mathrm { C }\) per second, where \(k _ { 1 }\) is a constant, until \(\theta = 60\). The heater then switches off.

1. Write down, in terms of \(k _ { 1 }\), how long it takes for the temperature to increase from \(40 ^ { \circ } \mathrm { C }\) to \(60 ^ { \circ } \mathrm { C }\). The temperature of the water then immediately starts to decrease at a variable rate \(k _ { 2 } ( \theta - 20 ) ^ { \circ } \mathrm { C }\) per second, where \(k _ { 2 }\) is a constant, until \(\theta = 40\).
2. Write down a differential equation to represent the situation as the temperature is decreasing.
3. Find the total length of time for the temperature to increase from \(40 ^ { \circ } \mathrm { C }\) to \(60 ^ { \circ } \mathrm { C }\) and then decrease to \(40 ^ { \circ } \mathrm { C }\). Give your answer in terms of \(k _ { 1 }\) and \(k _ { 2 }\). 4

7 Scientists can estimate the time elapsed since an animal died by measuring its body temperature.

1. Assuming the temperature goes down at a constant rate of 1.5 degrees Fahrenheit per hour, estimate how long it will take for the temperature to drop
   (A) from \(98 ^ { \circ } \mathrm { F }\) to \(89 ^ { \circ } \mathrm { F }\),
   (B) from \(98 ^ { \circ } \mathrm { F }\) to \(80 ^ { \circ } \mathrm { F }\). In practice, rate of temperature loss is not likely to be constant. A better model is provided by Newton's law of cooling, which states that the temperature \(\theta\) in degrees Fahrenheit \(t\) hours after death is given by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k \left( \theta - \theta _ { 0 } \right)$$ where \(\theta _ { 0 } { } ^ { \circ } \mathrm { F }\) is the air temperature and \(k\) is a constant.
2. Show by integration that the solution of this equation is \(\theta = \theta _ { 0 } + A \mathrm { e } ^ { - k t }\), where \(A\) is a constant. The value of \(\theta _ { 0 }\) is 50 , and the initial value of \(\theta\) is 98 . The initial rate of temperature loss is \(1.5 ^ { \circ } \mathrm { F }\) per hour.
3. Find \(A\), and show that \(k = 0.03125\).
4. Use this model to calculate how long it will take for the temperature to drop
   (A) from \(98 ^ { \circ } \mathrm { F }\) to \(89 ^ { \circ } \mathrm { F }\),
   (B) from \(98 ^ { \circ } \mathrm { F }\) to \(80 ^ { \circ } \mathrm { F }\).
5. Comment on the results obtained in parts (i) and (iv).

6 A hot drink is cooling. The temperature of the drink at time \(t\) minutes is \(T ^ { \circ } \mathrm { C }\).
The rate of decrease in temperature of the drink is proportional to \(( T - 20 )\).

1. Write down a differential equation to describe the temperature of the drink as a function of time.
2. When \(t = 0\), the temperature of the drink is \(90 ^ { \circ } \mathrm { C }\) and the temperature is decreasing at a rate of \(4.9 ^ { \circ } \mathrm { C }\) per minute. Determine how long it takes for the drink to cool from \(90 ^ { \circ } \mathrm { C }\) to \(40 ^ { \circ } \mathrm { C }\).

5. A bath is filled with hot water which is allowed to cool. The temperature of the water is \(\theta ^ { \circ } \mathrm { C }\) after cooling for \(t\) minutes and the temperature of the room is assumed to remain constant at \(20 ^ { \circ } \mathrm { C }\). Given that the rate at which the temperature of the water decreases is proportional to the difference in temperature between the water and the room,

1. write down a differential equation connecting \(\theta\) and \(t\). Given also that the temperature of the water is initially \(37 ^ { \circ } \mathrm { C }\) and that it is \(36 ^ { \circ } \mathrm { C }\) after cooling for four minutes,
2. find, to 3 significant figures, the temperature of the water after ten minutes. Advice suggests that the temperature of the water should be allowed to cool to \(33 ^ { \circ } \mathrm { C }\) before a child gets in.
3. Find, to the nearest second, how long a child should wait before getting into the bath.
   5. continued

1. The temperature, \(\theta ^ { \circ } \mathrm { C }\), of coffee in a cup, \(t\) minutes after the cup of coffee is put in a room, is modelled by the differential equation

$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 20 )$$ where \(k\) is a constant.
The coffee has an initial temperature of \(80 ^ { \circ } \mathrm { C }\) Using \(k = 0.1\)

1. use two iterations of the approximation formula \(\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { 0 } = \frac { y _ { 1 } - y _ { 0 } } { h }\) to estimate the temperature of the coffee 3 minutes after it was put in the room. The coffee in a different cup, which also had an initial temperature of \(80 ^ { \circ } \mathrm { C }\) when it was put in the room, cools more slowly.
2. Use this information to suggest how the value of \(k\) would need to be changed in the model.

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1. A teacher made a cup of coffee. The temperature \(\theta ^ { \circ } \mathrm { C }\) of the coffee, \(t\) minutes after it was made, is modelled by the differential equation

$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } + 0.05 ( \theta - 20 ) = 0$$ Given that

• the initial temperature of the coffee was \(95 ^ { \circ } \mathrm { C }\)
• the coffee can only be safely drunk when its temperature is below \(70 ^ { \circ } \mathrm { C }\)
• the teacher made the cup of coffee at 1.15 pm
• the teacher needs to be able to start drinking the coffee by 1.20 pm
  use two iterations of the approximation formula

$$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$$ to estimate whether the teacher will be able to start drinking the coffee at 1.20 pm .

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1. A cup of coffee is cooling down in a room. At time \(t\) minutes after the coffee is made, its temperature is \(x ^ { \circ } \mathrm { C }\), where $$x = 15 + 70 \mathrm { e } ^ { - \frac { t } { 40 } }$$
   1. Find the temperature of the coffee when it is made.
   2. Find the temperature of the coffee 30 minutes after it is made.
   3. Find how long it will take for the coffee to cool down to \(60 ^ { \circ } \mathrm { C }\).
2. 1. Use integration to solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 1 } { 40 } ( x - 15 ) , \quad x > 15$$ given that \(x = 85\) when \(t = 0\), expressing \(t\) in terms of \(x\).
   2. Hence show that \(x = 15 + 70 \mathrm { e } ^ { - \frac { t } { 40 } }\).

6 A cup of tea is served at \(80 ^ { \circ } \mathrm { C }\) in a room which is kept at a constant \(20 ^ { \circ } \mathrm { C }\). The temperature, \(T ^ { \circ } \mathrm { C }\), of the tea after \(t\) minutes can be modelled by assuming that the rate of change of \(T\) is proportional to the difference in temperature between the tea and the room.

1. Explain why the rate of change of the temperature in this model is given by \(\frac { \mathrm { d } T } { \mathrm {~d} t } = - k ( T - 20 )\), where \(k\) is a positive constant.
2. Show by integration that the temperature of the tea after \(t\) minutes is given by \(T = 20 + 60 \mathrm { e } ^ { - k t }\).
3. After 2 minutes the tea has cooled to \(60 ^ { \circ } \mathrm { C }\). Find the value of \(k\).

In an experiment a scientist considered the loss of mass of a collection of picked leaves. The mass \(M\) grams of a single leaf was measured at times \(t\) days after the leaf was picked. The scientist attempted to find a relationship between \(M\) and \(t\). In a preliminary model she assumed that the rate of loss of mass was proportional to the mass \(M\) grams of the leaf.

1. Write down a differential equation for the rate of change of mass of the leaf, using this model. [2]
2. Show, by differentiation, that \(M = 10(0.98)^t\) satisfies this differential equation. [2]

Further studies implied that the mass \(M\) grams of a certain leaf satisfied a modified differential equation $$10 \frac{dM}{dt} = -k(10M - 1), \quad (1)$$ where \(k\) is a positive constant and \(t \geq 0\). Given that the mass of this leaf at time \(t = 0\) is 10 grams, and that its mass at time \(t = 10\) is 8.5 grams,

1. solve the modified differential equation (1) to find the mass of this leaf at time \(t = 15\). [9]

Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20°C\). At time \(t\) minutes later, the temperature of the liquid is \(\theta°C\).

1. Explain how the information above leads to the differential equation $$\frac{d\theta}{dt} = -k(\theta - 20),$$ where \(k\) is a positive constant. [2]
2. The liquid is initially at a temperature of \(100°C\). It takes 5 minutes for the liquid to cool from \(100°C\) to \(68°C\). Show that $$\theta = 20 + 80e^{-(\frac{k}{5} \ln \frac{5}{3})t}.$$ [8]
3. Calculate how much longer it takes for the liquid to cool by a further \(32°C\). [3]

Scientists can estimate the time elapsed since an animal died by measuring its body temperature.

1. Assuming the temperature goes down at a constant rate of 1.5 degrees Fahrenheit per hour, estimate how long it will take for the temperature to drop
   1. from 98°F to 89°F,
   2. from 98°F to 80°F. [2]

In practice, rate of temperature loss is not likely to be constant. A better model is provided by Newton's law of cooling, which states that the temperature \(\theta\) in degrees Fahrenheit \(t\) hours after death is given by the differential equation $$\frac{d\theta}{dt} = -k(\theta - \theta_0),$$ where \(\theta_0\)°F is the air temperature and \(k\) is a constant.

1. Show by integration that the solution of this equation is \(\theta = \theta_0 + Ae^{-kt}\), where \(A\) is a constant. [5]

The value of \(\theta_0\) is 50, and the initial value of \(\theta\) is 98. The initial rate of temperature loss is 1.5°F per hour.

1. Find \(A\), and show that \(k = 0.03125\). [4]
2. Use this model to calculate how long it will take for the temperature to drop
   1. from 98°F to 89°F,
   2. from 98°F to 80°F. [5]
3. Comment on the results obtained in parts (i) and (iv). [1]