The temperature T(t) of a body at time t = 0 is 160^circ mathrmF and it decreases continuously as per the differential equation fracmathrmdTmathrmdt = -mathrmK(T - 80), where K is positive constant. If T(15) = 120^circmathrmF, then T(45) is equal to

Solution & Explanation

### Related Formula int fracdTT-T_s = -K int dt implies ln|T-T_s| = -Kt + C ### Core Logic Given fracdTdt = -K(T-80). Integrating from t=0 to t: int_160^T fracdTT-80 = -K int_0^t dt [ln|T-80|]_160^T = -Kt lnleft(fracT-8080right) = -Kt implies T(t) = 80 + 80e^-Kt Given T(15) = 120: 120 = 80 + 80e^-15K implies 40 = 80e^-15K implies e^-15K = frac12 To find T(45): T(45) = 80 + 80e^-45K = 80 + 80(e^-15K)^3 = 80 + 80left(frac12right)^3 = 80 + 80left(frac18right) = 90 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

Reference Study Guides

Differential Equations

More Differential Equations Questions — jee_main_2024_31_jan_evening

Practice all Differential Equations previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)