you place a cup of 200f coffee on a table in a room that is 67f and 10 minutes later it is 195f. approximately how long will it
be before the coffee is 180f
1 answer:
Newton's law of cooling says the rate of change of temperature is proportional to the difference between the object's temperature and the temperature of the environment.
Here, the object starts out at 200 °F, which is 133 °F greater than the environment temperature. 10 minutes later, the object is 195 °F, so is 128 °F greater than the environment. In other words, the temperature difference has decayed by a factor of 128/133 in 10 minutes.
The solution to the differential equation described by Newton's Law of Cooling can be written as the equation
T(t) = 67 + 133*(128/133)^(t/10)
where T is the object's temperature in °F and t is the time in minutes from when the object was placed in the 67 °F environment.
The equation
T(t) = 180
can be solved analytically, but it can be a bit easier to solve it graphically. A graphing calculator shows it takes 42.528 minutes for the temperature of the coffee to reach 180 °F.
Here, the object starts out at 200 °F, which is 133 °F greater than the environment temperature. 10 minutes later, the object is 195 °F, so is 128 °F greater than the environment. In other words, the temperature difference has decayed by a factor of 128/133 in 10 minutes.
The solution to the differential equation described by Newton's Law of Cooling can be written as the equation
T(t) = 67 + 133*(128/133)^(t/10)
where T is the object's temperature in °F and t is the time in minutes from when the object was placed in the 67 °F environment.
The equation
T(t) = 180
can be solved analytically, but it can be a bit easier to solve it graphically. A graphing calculator shows it takes 42.528 minutes for the temperature of the coffee to reach 180 °F.
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<u>FINAL ANSWER</u>
<u><em>In Decimal Form (Rounded to 3 significant figures)</em></u>
<em>x≈-1.24, 3.24</em>