According to Newton's law of cooling, the rate at which an object cools is directly proportional to the difference in temperature between the object and the surrounding medium. The face of a household iron cools from 135° to 107.5° in 30 minutes in a room that remains at a constant temperature of 80°. From calculus, the temperature f(t) of the face after t hours of cooling is given by f(x) = 55(2)^(−2t) + 80.
Assuming t = 0 corresponds to 1:00 P.M., approximate to the nearest tenth of a degree the temperature of the face at 2:00 P.M., 3:30 P.M., and 4:00 P.M.
Ah, Newton's law of cooling, the perfect equation for when you want to sound cool while talking about cooling. Alright, let's crunch some numbers and find out the approximate temperatures.
At 2:00 P.M., we have 1 hour of cooling. Plugging in t = 1 into the equation f(t) = 55(2)^(-2t) + 80, we get:
f(1) = 55(2)^(-2*1) + 80
= 55(2)^(-2) + 80
= 55(1/4) + 80
= 13.75 + 80
= 93.75
So, at 2:00 P.M., the temperature of the iron's face would be approximately 93.8 degrees.
Now, moving on to 3:30 P.M., we have 2.5 hours of cooling. Plugging in t = 2.5 into the equation, we get:
f(2.5) = 55(2)^(-2*2.5) + 80
= 55(2)^(-5) + 80
= 55(1/32) + 80
= 1.71875 + 80
= 81.7
So, at 3:30 P.M., the temperature of the iron's face would be approximately 81.7 degrees.
Lastly, at 4:00 P.M., we have 3 hours of cooling. Plugging in t = 3 into the equation, we get:
f(3) = 55(2)^(-2*3) + 80
= 55(2)^(-6) + 80
= 55(1/64) + 80
= 0.859375 + 80
= 80.9
So, at 4:00 P.M., the temperature of the iron's face would be approximately 80.9 degrees.
Remember, these are just approximate values, so don't get too heated if they're not exact!
To approximate the temperature of the face at 2:00 P.M., 3:30 P.M., and 4:00 P.M., we need to convert the given times to hours and substitute them into the temperature equation.
To convert the time to hours, we can use the fact that t = 0 corresponds to 1:00 P.M.:
1 hour = 1:00 P.M. to 2:00 P.M.
2.5 hours = 1:00 P.M. to 3:30 P.M.
3 hours = 1:00 P.M. to 4:00 P.M.
Now, let's substitute the values into the temperature equation f(t) = 55(2)^(-2t) + 80 and calculate the temperature at each time:
1. For 2:00 P.M. (1 hour):
f(1) = 55(2)^(-2 * 1) + 80
= 55(2)^(-2) + 80
= 55(1/4) + 80
= 13.75 + 80
= 93.75
So, the approximate temperature at 2:00 P.M. is 93.75°.
2. For 3:30 P.M. (2.5 hours):
f(2.5) = 55(2)^(-2 * 2.5) + 80
= 55(2)^(-5) + 80
= 55(1/32) + 80
= 1.71875 + 80
= 81.71875
So, the approximate temperature at 3:30 P.M. is 81.7°.
3. For 4:00 P.M. (3 hours):
f(3) = 55(2)^(-2 * 3) + 80
= 55(2)^(-6) + 80
= 55(1/64) + 80
= 0.859375 + 80
= 80.859375
So, the approximate temperature at 4:00 P.M. is 80.9°.
Therefore, the temperature at 2:00 P.M., 3:30 P.M., and 4:00 P.M. are approximately 93.8°, 81.7°, and 80.9°, respectively.
To approximate the temperature of the face at different times, we need to substitute the given times into the equation f(t) = 55(2)^(-2t) + 80, where t is the time in hours since 1:00 P.M.
To find the value of t for each time, we need to calculate the time difference between each given time and 1:00 P.M.
1. 2:00 P.M. - 1:00 P.M. = 1 hour
So, t = 1
Substituting t = 1 into the equation:
f(1) = 55(2)^(-2(1)) + 80
= 55(2)^(-2) + 80
= 55(1/4) + 80
= 13.75 + 80
≈ 93.8°F
Therefore, the temperature of the face at 2:00 P.M. is approximately 93.8°F.
2. 3:30 P.M. - 1:00 P.M. = 2.5 hours
So, t = 2.5
Substituting t = 2.5 into the equation:
f(2.5) = 55(2)^(-2(2.5)) + 80
= 55(2)^(-5) + 80
= 55(1/32) + 80
= 1.71875 + 80
≈ 81.7°F
Therefore, the temperature of the face at 3:30 P.M. is approximately 81.7°F.
3. 4:00 P.M. - 1:00 P.M. = 3 hours
So, t = 3
Substituting t = 3 into the equation:
f(3) = 55(2)^(-2(3)) + 80
= 55(2)^(-6) + 80
= 55(1/64) + 80
= 0.859375 + 80
≈ 80.9°F
Therefore, the temperature of the face at 4:00 P.M. is approximately 80.9°F.
f(t) = 55 (2)^(-2t) + 80
now you tell me that if it starts at 135 it will be at 107.5 in 0.5 hours
let me check that
yes, it is 135 at t = 0
now at t = 0.5
it is
55 (2)^(-1) + 80
= 27.5 + 80
= 107.5 sure enough I believe you
now the problem
say at 2 pm..... then t = 1 hour after 1 pm
so
f(t) = 55 (2)^(-2t) + 80
= 55 * 2^-2 + 80
= 55/4 + 80
= 93.75 at 2 pm
then use t = 2.5 and t = 3 for 3:30 and 4