Question 1165586

According to Newton’s law of cooling, the temperature of an object changes at a rate proportional to the difference in temperature between the object and the outside medium. If an object whose temperature is 70OF is placed in a medium whose temperature is 20O and is found to be 40O after 3 minutes, what will its temperature be after 6 minutes?
a. 25OF b. 28OF c. 31OF d. 34OF
<pre>Newton's law of cooling formula: {{{matrix(1,3, T(t), "=", T[s] + (T[o]  -  T[s]) * e^(- kt))}}} where: {{{t}}} = time (t) at a COOLED temperature (3 minutes, in this case)
                                                                    {{{T(t)}}} = TEMPERATURE (T) at a given time (t) (400<sup>o</sup>F, in this case)
                                                                    {{{T[s]}}} = SURROUNDING temperature (200<sup>o</sup>F, in this case)
                                                                    {{{T[o]}}} = ORIGINAL/INITIAL temperature (700<sup>o</sup>F, in this case)
                                                                    {{{k}}} = CONSTANT or COOLING rate (Unknown, in this case)


     In this case, we 1st have to determine the value of k, and so: {{{matrix(8,3, T(t), "=", T[s] + (T[o]  -  T[s]) * e^(- kt),
400, "=", 200 + (700 - 200) * e^(- 3k),
400, "=", 200 + 500e^(- 3k),
400 - 200, "=", 500e^(- 3k),
200/500, "=", e^(- 3k),
2/5, "=", e^(- 3k), 
- 3k, "=", ln (2/5),
k, "=", ln (2/5)/(- 3))}}}

{{{matrix(1,3, T(t), "=", T[s] + (T[o]  -  T[s]) * e^(- kt))}}}, where: {{{t}}} = time (t) at a COOLED temperature (6 minutes, in this case)
                                    {{{T(t)}}} = TEMPERATURE (T) at a given time (t) (Unknown, in this case)
                                    {{{T[s]}}} = SURROUNDING temperature (200<sup>o</sup>F, in this case)
                                    {{{T[o]}}} = ORIGINAL/INITIAL temperature (700<sup>o</sup>F, in this case)
                                    {{{k}}} = CONSTANT or COOLING rate ({{{ln (2/5)/(- 3)}}}, in this case)

                                    {{{matrix(2,3, T(6), "=", matrix(2,1, "", 200 + (700 - 200) * e^(- 6 * ((ln (2/5))/(- 3)))), T(6), "=", matrix(2,1, "", 200 + 500e^(- 6 * ((ln (2/5))/(- 3)))))}}}
   Temperature, 6 minutes after, or {{{highlight_green(matrix(1,6, T(6), "=", 280^o, F, "(CHOICE", "b.)"))}}}</pre>