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Lubov Fominskaja [6]
2 years ago
15

Suppose you cool a pot of soup in a 20℃ room. Right when you take the soup off the stove, you measure its temperature to be 180℃

. After 20 minutes, the soup has cooled to 100℃. Suppose you can eat the soup when it is 80℃. How long will it take to cool to this temperature? What will be the temperature of the soup after 45 minutes?
Mathematics
1 answer:
lara31 [8.8K]2 years ago
3 0

Answer:

it will be cool to 220℅because heat is increase by 100degrees in every after 20 minutes

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Answer:

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Step-by-step explanation:

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Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

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Let the amount salt in the tank at any time t be Q(t).

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Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

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In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

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\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

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