Question

Two large containers A and B of the same size are filled with different fluids. The fluids in containers A and B are maintained at 0° C and 100° C, respectively. A small metal bar, whose initial temperature is 100° C, is lowered into container A. After 1 minute the temperature of the bar is 90° C. After 2 minutes (since being lowered into container A) the bar is removed and instantly transferred into the other container. After 1 minute in container B the temperature of the bar rises 10°. How long, measured from the start of the entire process, will it take the bar to reach 99.5° C? (Round your answer to two decimal places. Assume the final temperature being asked for is reached while the bar is container B.)

t = ______ min

Answer 1

Ô solution : Z Given that Container A has a Temperature of OC and containeo B hao a Temperatute of 100°C Foomula : - nedton's lod of cooling is given by the Equation dt -=K(T-Tm) dt Ks constant of popokonality I is the terperalue of the observed object This the ambical Temperature dt (T-Tm) = kot Si dr = Sudé Det ska - ] In IT-Tml = kt+C T-Im=COKE T=cert + Tm. Fool We will consideo when the bar is immezsed in conta. inco A. we know that the initiale Temperature of the bar is 1006 al t=0.

that al we know The ambient temperature .foo this case is Tm=0. The Tempecabute of contineo A, finally t: I, The bari's Temperature is 90°6. using the above reformation, we can solve Equalian O Foo contairco A. 100-ce +0 Ti€:13 G-100 ке 50 loe have T = looe to o To solve fook We will use Z=1 and T=90° (K) -90-100C 98 100 : K 10 [p. ext]. en 19k K value in Equation @ englot T:1002 T-2013) given T-9950 99.6°100) .995- (78) 99.69 62) 100

en[.993)-tln() ki In10.995) en 1966 t: (-50125#100) -0-105360515 % =0.047375 minutes The bat has a temperature of 99.5% neat, we consides when the bar ig immerged in i containco B. t: 2 2 2013) T= 100 (ie T 10018 100) T-81 to! T:81. t. 1, T-81710-91. and in 100 lifoom ago) Les 81-ce+100 81-100 -601) [:en] C=-19 "C value in Equaliona T =-1geht +100

6:12 โ: 91 solve Foo K 91-100=-elok +9=41acht 91:-190 xBf 100 olo cekti) en la alk 56. The Equalion modeling the temperature of the bar is containco B is en alt T= -190 + 100 3 T= 99 og in En 99.5° -jalnost 7100 99.3-100= -1966024 ent +0.5=7196 Oodalit -C 19 en105) en (alt 13•63758616 = (10. 707214401) Z t = 3•63758616 0.747214401 -140868196 t: 4.868196+ 0.047375 4915770 | t: 00047575 min t-46915770

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