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Marrrta [24]
3 years ago
15

In the two-stage cooling method, what is the maximum amount of time allowed to cool food from 135 degrees F or more to 70 degree

s F or less?
Physics
1 answer:
Tresset [83]3 years ago
5 0
The problem ask to find the maximum amount of time that is allowed to cool food from 135 degree F or more to 70 degree F or less and the allowed time would be that the food would be cool of is around to 4 to 6 hours. I hope you are satisfied with my answer and feel free to ask for more of you have questions and further clarifications 
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The specific heats of aluminum and iron are 0.214 and 0.107 calories per gram degrees Celsius​ [cal/(g degrees Upper C ​)] respe
tester [92]

Answer:

Answer for question "

The specific heats of aluminum and iron are 0.214 and 0.107 calories per gram degrees Celsius​ [cal/(g degrees Upper C ​)] respectively. If we add the same amount of energy to a cube of each material​ (of the same​ mass) and find that the temperature of the aluminum increases by 28 degrees Fahrenheit ​[degrees Upper F ​], how much will the iron temperature increase in degrees Fahrenheit ​[degrees Upper F ​]? "

is explained in attachment.

Explanation:

Download pdf
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3 years ago
Which example is not a part of the electromagnetic spectrum
dsp73
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2 years ago
Which scenarios are examples of physical change?​
Oxana [17]

Answer:

For example: Freezing, boiling, are physical

Explanation:

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What would you use to measure distance east and west of the prime meridian
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7 0
3 years ago
What is the equivalent capacitance of the three capacitors in the figure (figure 1)?
vekshin1

The equivalent capacitance of the combination is  \dfrac{1}{C} = \dfrac{1}{C_1}+\dfrac{1}{C_2}  where C1 and C2 are the capacitance of both capacitors in series.

<h3>What is equivalent capacitor?</h3>

Let the capacitance of both capacitors be C1 and C2. For a series connected capacitors, same charge flows through the capacitors but different voltage flows through them.

Let Q be the amount of charge in each capacitors,

V be the voltage across each capacitors

C be the capacitance of the capacitor.

Using the formula Q = CV where V = Q/C... (1)

For the large capacitor with capacitance of the capacitor C1,

Q = C_1V_1;

V_1 = \dfrac{Q}{C_1}...(2)

where V_1 is the voltage across C_1,

For the small capacitor with capacitance of the capacitor C_2,

Q = C_2V_2;

V_2 = \dfrac{Q}{C_2} ... (3)

where V_2 is the voltage across C_2,

Total voltage V in the circuit will be;

V = V_1+V_2... (4)

Substituting equation 1, 2 and 3 in equation 4, we have;

\dfrac{Q}{C} = \dfrac{Q}{C_1} +\dfrac{ Q}{C_2}

\dfrac{Q}{C} = Q({\dfrac{1}{C_1}+\dfrac{1}{C_2})

Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;

\dfrac{1}{C} = ({\dfrac{1}{C_1}+\dfrac{1}{C_2})

This gives the equivalent capacitance of the combination.

To know more about equivalence capacitance follow

brainly.com/question/5626146

#SPJ4

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