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The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature

Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

$T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$

Where

$(t) = \ time$

$T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)$

$T_{s} = Surrounding \ temperature$

$T_{0} = Initial \ temperature \ of \ the \ body$

$k = constant$

From the question,

$T_{0} = 86 ^{o}C$

$T_{s} = -20 ^{o}C$

∴ $T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C$

$T_{0} - T_{s} = 106^{o} C$

Therefore, the equation $T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}$ becomes

$T_{(t)}=-20+106 e^{kt}$

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time $t = 1 \ hour$ , $T_{(t)} = 58^{o}C$

Then, we can write that

$T_{(1)}=58 = -20+106 e^{k(1)}$

Then, we get

$58 = -20+106 e^{k(1)}$

Now, solve for $k$

First collect like terms

$58 +20 = 106 e^{k}$

$78 =106 e^{k}$

Then,

$e^{k} = \frac{78}{106}$

$e^{k} = 0.735849$

Now, take the natural log of both sides

$ln(e^{k}) =ln( 0.735849)$

$k = -0.30673$

This is the value of the constant $k$

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at $t = 2 \ hours$

From

$T_{(t)}=-20+106 e^{kt}$

Then

$T_{(2)}=-20+106 e^{(-0.30673 \times 2)}$

$T_{(2)}=-20+106 e^{-0.61346}$

$T_{(2)}=-20+106\times 0.5414741237$

$T_{(2)}=-20+57.396257$

$T_{(2)}=37.396257 \ ^{o}C$

$T_{(2)} \approxeq 37.40 \ ^{o}C$

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

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A square has an area that is less than 100 m. What is a reasonable range for the graph of the

WITCHER [35]

1-9 I’m pretty sure but idk

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All of the following are rational except _____. - 3.14159... 0

sergejj [24]

Answer:

All of the following are rational except 3.14159...

Step-by-step explanation:

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the diagram, JG = 5 cm and GE = 10 cm. Based on this information, can G be a centroid of triangle HJK?

SIZIF [17.4K]

Point G cannot be a centroid because GE is wider that JG or JG is shorter than GE. So in this diagram GE is wider than JG with 10 cm and 5 cm respectively based on this information Point G cannot be a centroid of triangle HJK. So the answer is point G cannot be a centroid because JG is shorter than GE.

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3 years ago

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An airplane ascends such that its gain h in altitude is proportional to the square root of the change x in horizontal distance t

Mazyrski [523]

Answer:velocity=185 m/s

Step-by-step explanation:

So make k the subject of the formula by divide both by sqrt(x)

h= k × sqrt(x)

k=h/sqrt(x)

k=256 / sqrt(256)

k=256 / 16

k=16

So now substitute the value of k :

h= 16 × sqrt (x)

Then differentiate:

=(1/2 × sqrt(x))× 16

=16/(2×sqrt(x)

=8/sqrt(x)

Then

= 8/sqrt(x) × 370

=8/ sqrt(256) × 370

=8/16 × 370

velocity=185 m/s

6 0

3 years ago