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 0C and 100C, respectively. A small metal bar, whose initial temperature is 100C, is lowered into container A. After 1 minute the temperature of the bar is 90C. After 2 minutes (since being lowered into container A) the bar is removed and instantly transferred to the other container. After 1 minute in container B, the temperature of the bar rises 10C. How long, measured from the start of the entire process, will it take the bar to reach 99.7 C? (Round your answer to 2 decimal places. Assume the final temperature being asked for is reached while the bar is container B.)

t=______min

Answer 1

We know that di = R(I-T6) dt integrating con bothsides, we get dT - kfdt T-T5) 7 In (T-Tn) = kt te T-T5 = elekt 키) Ub 1 • Here, T= 10000 Tr = 0°c and time - 0 min subal these values in above eq (i), we dubaliating get c=100 1 and if to 1 min temperature of the bar T= 90° ettre lubiłbuting gium 90°C = (100) e slata in abone eq, we get K. (0) ek Oog K = ln (0.9) Ko - 0.1053 •If t=2 min, (-0.1053)(2) P-0 = (100) e T = (100) (0.81) T= 8106 Now, 81 = 100+ Ce kilo) C--19 CSScanned with CamScanner

10 ich temperature is toc , then the new Temperature T= 81+10 - 90°C KOU) 91 = 100+ (-19) e ek = 9/19 K = lnl en 1970 K= 0.7472 therefore, e 99.7 (-0.7472)60 = 100 + (-19) e (-0.7472) (+) -0.3 19 therefor time to en (003 19 (0.7472) - 401484 -0-7472 t = 5055 min therefore, after 2 min of time, then we get t=2+.5.55 t= 7.55 min CS$cahned with CamScanner