Question

(D) 70 m TUL 20. An object is cooling off according to Newton's law of cooling. Initially it is at 70°C. 5.0 minutes later it is at 50°C. 5.0 minutes after that it is at 38° C. What is the temperature of the surroundings? uoy bitovelvol 92 9vode ist woh (A) (D) 16°C 28°C (C) (B) (E) 20°C 32° C 24°C sol el SI (0) 315 vod dort ni 1911 or mort also boold til siege of 21 Jaibiydgoid A al

Answer 1

- initial temperature To = 70 c. temperature after 5 minutes - Tlt) = 500 so time t - 5 minutes temperature after next 5 miscites I(t) = 38'C So tz – 5 minutes since we know that from the Newton's law of cooling- T(t) = Ts & (To - Ts) e akt where T() = temperature after timet To = Sunnounding temperature To = initial temperature k= constant 80 for first five minutes - T (t) = Ts +(To-Ts) ēkt, putting the values 50 = To +(70 - Ts) e-k(5) 50-To - 70-Ts) e-k(5) - e-k(5) 50-TS Y 770 - T -

for next 5 minutes - Te(t) = Ts + (T, (t) - Ts) ēkt z putting the values - 38 = Tz + (50-To) e-kis) & k(5) = 38-Is I 150-To Now from eg. 0 and eq. ③ - (38 - Is I - 150 - Ts 1 (5o-T) 1 70-Ts ) → (38–1] (70-1) = (50 - La }} → 2660 - 381, -707 +1= 2500 +1° -10075 > 2660 – 2500 = -100 Is + 108 Is >> Ts = too ★ To - 20:00 so thus the temperature of surrounding is 20°c option (B) is right option.