The temperature T t of a body at time t = 0 is 160 ° F and it decreases continuously as per the differential equation d T d t = − K T − 80 , where K is positive constant. If T 15 = 120 ° F , then T 45 is equal to

Step-by-step solution

Given: d T d t = − K T − 80 ⇒ d T T − 80 = − K d t ⇒ ∫ 160 T d T T − 80 = ∫ 0 t − K d t ⇒ log T − 80 160 T = − K t ⇒ log T − 80 − log 80 = − K t ⇒ log T − 80 80 = − K t ⇒ T = 80 + 80 e − K t Now, using the value T 15 = 120 ° we get, ⇒ 120 = 80 + 80 e − K · 15 ⇒ 40 80 = e − 15 k ⇒ e − 15 k = 1 2 ∴ T 45 = 80 + 80 e − 45 k ⇒ T 45 = 80 + 80 e − 15 k 3 ⇒ T 45 = 80 + 80 × 1 8 ⇒ T 45 = 90 ° F