Unformatted text preview: H ( t ) = 2 × 2 = 4 ◦ ± per hr . We are given that T (0) = 110, and that the temperature M ( t ) outside the tank is a constant 80 ◦ ±. Hence the temperature in the tank is governed by dT dt = 1 64 [80 − T ( t )] + 4 = − 1 64 T ( t ) + 5 . 25 , T (0) = 110 . Solving this separable equation gives T ( t ) = 336 + Ce − t/ 64 . To Fnd C , we use the initial condition to see that T (0) = 110 = 336 + C ⇒ C = − 226 . This yields the equation T ( t ) = 336 − 226 e − t/ 64 . 122...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Constant of integration, Tank, 1000 btu, 0.2185 35.83 30.4 min, 2000 Btu

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