Question

2. (4 pts) A small metal bar, whosetemperature was 20 C at 2 pm was dropped into a container of boiling water. After 10 minutes its temperature rose to 25 C. The metal bar was left in the water until 2:20pm. (a) What was its temperature when it was taken out of the water at 2:20pm? (b) Suppose the bar was then put in in a refrigerator which had a fixed temperature of OC at 2:20pm. At 2:30pm, the temperature of the bar was 15 C. Suppose the bar stayed in the refrigerator. What was its temperature at 3pm? (Comment: basically you are asked to solve two heating/cooling problems. The initial condition for (b) relies on your answer in (a).

Answer 1

Solvtipn.- Given that a) Rate of change of with Time = ? at 2pm Temperature of bar was 20°c after 10 min i.e; at 2:10pm temp. of bar becomes 25°c. 5o, Rate of change of Temp. with time de 85.600.65 o/min. so, at 2:20pm. Temp. of bar will be W 2010 -) T2-20510 dI - llac min Ta = ? to-t = 2:20 pm - 2:00 Pm s domin b) By Newtons Law of cooling I - - *(T-ta) solucion of above differential equation will be T(t) = Tat (To - Tajekt S K= constant To = initial Temperature Ta = o(surrounding is fridge which is at oºc.) To = 30°c

so = 300 T = Temperature at a time Ta = Temp. of surrounding: here Ta=oc Ilt)= 0+ (30-ojekt I(t)= 30€ kt. At 2:30pm is ine, after 10 min temp: 15° as per question ie; T (t) = 15 t= 10 min. 15= 300kcio) Yaz ēlok logiya)= -10% log (1) - dogeca)- -10% 0-.6930=-10K K=.069 | Temp. at 3:pm ie; after 40 min temp = ? T(40) = 30 ē (0.069) x40. -30 2.76. 51.89 a 2°C] 2cm . Thank you,