Newton's Law of Cooling on cup of coffee

[ProBasket]

A cup of coffee at 174 degrees is poured into a mug and left in a room at 76 degrees. After 5 minutes, the coffee is 134 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt = k(T-76)

here's what i done:

y(0) = 98e^{kt}

y(5) = 134-76 = 58, so...

98e^{5k} = 58
k = -0.104904
y(15) = 98*e^{15*-0.104904} + 76
and got 159.8448 which is wrong. anyone know where i went wrong?

 

[Integral]

Go back to the general solution of the DE, you have made some mistakes in how you have handled the constants.

 

[xanthym]

A cup of coffee at 174 degrees is poured into a mug and left in a room at 76 degrees. After 5 minutes, the coffee is 134 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt = k(T-76)

here's what i done:

y(0) = 98e^{kt}

y(5) = 134-76 = 58, so...

98e^{5k} = 58
k = -0.104904
\ \ \ \color{red} y(15) = 98*e^{15*-0.104904} + 76
and got 159.8448 which is wrong. anyone know where i went wrong?

All your work is basically correct (although the presentation could improve a bit). You've made a careless mistake computing the final equation (in RED above). That equation is correct and evaluates to:
{Temperature at t=(15 min)} = (96.316 deg)


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