Derivatives and exponential function Part 1&2 fairly accurate model to describe

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Derivatives and exponential function Part 1&2

fairly accurate model to describe the temperature changes in a conducting object is Newton's Law of Cooling. Suppose that at a time t is great than or equal to 0, an object has a temperature of T(t). The Law of Cooling says that the rate at which the temperature of the object increases or decreases i given by dT k(T(t-A) (1) dt here A is the ambient (surrounding) temperature and k 0 is a constant called the conductivity (which is a property of the cooling object). Newton's Law of Cooling assumes that the cooling body has a uniform temp throughout its interior. This is not strictly accurate, as a cooling body looses heat through its surface. 1. Explain in words what equation 1 means. Spefically, in terms of T and A, when is dTIdt 0 and when is dTVdtKO? For the case of hot coffee cooling to room temperaute, which case do you expect to See 2. Verify by subsitutuion that the solution to equation 1 subject t the initial condition T(0) Ta is (2) T (t) A+(T A)e 4. Consider the case of a hot cup of coffee cooling with an ambient room temperature of AE60degrees F and the initial temp of the coffee is Ta 200degreeF. Use a graphing utility to plot the temp function for k .3, .2, .2 and .05. Comment on how the curves change with k. Do larger values of k produce faster or slower rates of temp change? 5. For the values of A and To in question 4, estimate the value of k that describes the case in which the coffee cools to 100degrees in 10mins. Next, suppose you want to cool your hot coffee to 100degrees F as quickley as possible. Suppose also that you have one ounce of cold milk with a temp of 40degreeF that you can add to the cooling coffee at any point. When should you add the milk to cool the coffee to 100degreeF as quickley as possible? 6. We need to make an assumption about the effect ofcold milk on the temp. of the coffee. A resonable assumption is that when milk is added to coffee, the temp of the coffee immediately decreases to the average of the coffee temp and the milk temp, where the average is weighted by the volumes. So if we add 1 oz of milk with temp T to 8 oz with temp T, the temp of the mixture wi be 1 X Tm 8 XT Tim 8T (3) 1 +8 (For example, if we add 1 oz of 40degreesF milk to 8 oz of 150degreeF coffee, the coffe temp is lowered immediatly to about 138degreeF.) Assume the coffe is alloed to cool fot ti mins, here t be determined. Use equation 2 and the value of k found in question 5 to determine the temp of the coffee at t-t Use assumption 3 to determine the temp of the coffee when 1 oz of 40degreeF milk is added to 8 oz of coffee, which has a temp of T. call this new coffee temp Tnew 8. Use equation 2 to show that the temp of the coffee for t t1 using Tnew as te initial temp is here t is now measured in ins after T9t) 60+(Tnew-60) e 9. We now solve for the value of t such that T(t) 100 and call this value t2. Remember that t2 measures time after the milk is added. Show that 1In 10. Notice that as 1 increases, t2 decreases. In fact, show that t1 >8.65 (approximately then t2 0 and the solutions are not meaningful. Therefore, we consider t1 in the interval 0

Explanation / Answer

1)

Temperature difference in any situation results from energy flow into a system or energy flow from a system to surroundings. The former leads to heating, whereas latter leads to cooling of an object.
Newton

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