Question

Quiz: 4.6 QUIZ Time Remaining: 01:47:19 Submit Quiz This Question: 1 pt 5 of 5 (4 complete) ID This Quiz: 5 pts possible Estabon poured himself a hot beverage that had a temperature of 200°F then set it on the kitchen table to cool. The temperature of the kitchen was a constant 71°F. If the drink cooled to 185 °F in 4 minutes, how long will it take for the drink to cool to 98°F? minutes (Do not round until the final answer. Then round to the nearest minute as needed.)

Answer 1

By the cooling lame of Membon a = k (Ts-T) O of Thermodynamics. dT at de change in temperation of body with respect to time ka constant T = Surrounding temperations TL = Body temperature at time it! Given es, Kitchen temperature (surrounding temperature) Ts = 70F initial temperators of Body T = 200°F and body reached at los °F in 4 minuty. then have much time it take to reach 90°F. By G.D d (T2-T) K (Ts-To) at dI = at T2210SF 07 K (Ts Tal t24 minute Ik alt J (Ts-T2) Ezo. - 200P 10S 4 minute Ilog (Ts-72)7 toge = (kt). {log (Ts-105) – bg (75-2003 top: 4x60k realme c2ay ( ) + = 240K

Ts=70F no. by (o) = SUOL lg(1) = quoka jka 240 rby (2) from cooling 200 F to 90'F C And now be dy is to dI = kltsto) Lot = kot at T-T2 -T2 2.900 itzt (kt) Is-T2 ): tzo . T2 = 206F - 90 by s 200 ) = kt Ts a 200 by [71-18) = kt by (92) = Le loy changes t' dost' = 2008, by (224) .240 seconds leg (1 5 a t'= 240x 12-65 see ť= 3036. 50 second 36 secondo sominate or It= from to cart a Sedly So, the time tabe 50 minule 3 Orsen it takes time 200°F to 100€ realme C2