41. Cooling soup Suppose that a cup of soup cooled from 90°C t...

Question

41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.

a. How much longer would it take the soup to cool to 35°C?

Accepted Answer

Welcome back everyone. A metal rod is heated to 120 °C and cools to 80 °C in 5 minutes in a room at 40 °C. According to Newton's law of cooling, how much more time will it take to reach 50 °C? A 20 minutes, B 10 minutes. C 25 minutes, and D 15 minutes. So for this problem, let's remember Newton's law of cooling. Specifically, temperature at time T is equal to. T subscripts, the surrounding temperature or the ambient temperature, plus T subscript 0, which is the initial temperature minus T subscripts. Multiplied by e to the power of negative kt, right, where k is a positive constant representing the cooling rate. So first of all we will find the cooling constant k where given that after 5 minutes the rod cools to 80 °C, so we can set 80. Equals CS, right? That would be. The surrounding temperature 40 plus. T subscript 0 is. 120. Degrees Celsius, that's the initial temperature. Minus Cs, so -40 multiplied by e to the power of negative k multiplied by C, right? And C in this case is 5. So now let's solve for K. Let's subtract 40 from both sides, so that 40 is equal to 80. E to the power of -5 k. Now we can divide both sides by 80 to get one half equals e to the power of -5 k. And now we can take the natural logarithm of both sides, ln of 1/2, which can also be written as negative aln of 2, right. is equal to -5k, so that k is equal to ln of 2 divided by 5. Well done. So now that we have our constant k, we will find the total time to reach 50 °C. We will solve for the time it takes to cool from the initial 120 °C down to 50 °C. So now we're setting 50 equals again 40 plus 120 minus 40. Multiplied by e to the power of. Now, let's write negative K multiplied by T total. Again, let's subtract 40 from both sides so that we get some. Equals 80 e. So the power of negative KT total, total time, right? And now we're going to divide both sides by 80, which gives us 1/8. Equals e to the power of negative kc total. And we can take a land of both sides. So that negative eon of 8 equals. Negative K multiplied by T total. And now let's substitute the value of K, right? So that'd be negative of 2 divided by 5 multiplied by T total. And from here we can essentially find the total time. Let's cancel out the negative signs, multiply both sides by 5, and divide it by eon of 2, so that the total time. Is equal to 5 multiplied by eon of 8. Divided by ln of. 2 And this is equal to. 15 minutes, right? So now, what is the additional time? Well, the question asks how much more time it will take. For the 8080 °C mark, so the additional time is going to be the total time minus 5 minutes given to us. So the additional time we can call it delta T is equal to 15 minutes. The total time minus 5 minutes, which is 10 minutes. And the correct answer is option B. Thank you for watching.

41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.a. How much longer would it take the soup to cool to 35°C?

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41. Cooling soup Suppose that a cup of soup cooled from 90°C to 60°C after 10 min in a room where the temperature was 20°C. Use Newton’s Law of Cooling to answer the following questions.
a. How much longer would it take the soup to cool to 35°C?