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College Physics

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Expert Solution

Step 1

Based on the Newton's law of cooling the temperature is related with time as:

$T = T_{a} + (T_{0} - T_{a}) e^{- k t}$ ...1

$T_{a} = T h e t e m p e r a t u r e s u r r o u n d i n g t h e o b j e c t T_{0} = t h e i n i t i a l t e m p e r a t u r e o f t h e o b j e c t t = t h e t i m e i n m i n u t e s T = t e m p e r a t u r e o f t h e o b j e c t a t t i m e t k = D e c a y c o n s \tan t$

Given:

$T_{a} = T h e t e m p e r a t u r e r e f r i g e r a t o r = 38^{o} F T_{0} = t h e i n i t i a l t e m p e r a t u r e o f t h e s o d a c a n = 67^{o} F t = t h e t i m e i n m i n u t e s = 30 m i n T = t e m p e r a t u r e o f t h e s o d a c a n a t t i m e 30 m i n = 58^{o} F k = D e c a y c o n s \tan t$

Now putting the values in equation 1

$58^{o} F = 38^{o} F + (67^{o} F - 38^{o} F) e^{- k (30 m i n)} \frac{20^{o} F}{29^{o} F} = e^{- k (30 m i n)}$

Now taking log we get

$\ln (0.69) = - k 30 k = 0.012$

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