Question

Differential equations

8. (9 points) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. An object initially at 25°C is put outside, where the temperature is 0°C. If the temperature of the object has dropped to 15°C after 20 minutes, determine when will it drop to 5°C ?

Answer 1

dT/dt alpha alpha (T_0 - T_s) dT/dt = alpha (T - T_s) dT/T - Ts = alpha dt ln (T - Ts) = alpha t + e when t = 0, T = 25, T_s = 0 therefore e = ln 25 when t = 20, t = 15 therefore ln (15) = alpha times 20 + ln 25 ln (15) - ln 25/20 = alpha ln (t - t_s) = ln (15/25) times t/20 + ln 25, Since T_s = 0 ln(T) = ln(15/25) times t/20 + ln 25 (ln 5/25) = ln (3/5) times t/20 t = ln (1/5)/ln(3/5) times 20 = -1.6094/-0.5108 times 20 = 63. 0132 after t = 63.0132 minute temperature will drow at 6 degree e