The temperature on Monday in Minot, North Dakota was 3°F. On Tuesday, the temperature
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86.34 minutes was used to cool the soup to 80 °F
<h3></h3><h3>How can we calculate this using Newton's Law of Cooling?</h3>Newton's Law of Cooling can be computed as :
The initial temperature is been represent by T0 which is ( 100 F)
The temperature of surrounding is been given as (69 F)
The final temperature is been given as T
t = time
we were told that the temperature of the soup move to 95 F, using 15 minutes.
Hence, T = 95 F
t = 15 min
Then we can input the values into the above equation as :
if we simplify this , we have
Then
Then we can find the value of k as :
k= -0.012
Then we have the new equation as
Then at T=80F
we have,
t=86.34
Then if we simplify , we have t=86.34 which implies that 86.34 minutes was used to cool the soup to 80 °F
Learn more about Newton's Law of Cooling at:
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7v = 2v - 20
First, subtract 2v from both sides. / Your problem should look like: 7v - 2v = -20
Second, simplify 7v - 2v to 5v. / Your problem should look like: 5v = -20
Third, divide both sides by 5. / Your problem should look like: v =
Fourth, simplify
to 4. / Your problem should look like: v = -4
Answer: v = -4
First, subtract 2v from both sides. / Your problem should look like: 7v - 2v = -20
Second, simplify 7v - 2v to 5v. / Your problem should look like: 5v = -20
Third, divide both sides by 5. / Your problem should look like: v =
Fourth, simplify
Answer: v = -4