Isaac's law says,
T(t) = Ta + (Ts - Ta)e^kt
where Ta is the room temp
Ts is the temp of the object
and t is the time
T(6) = 72 + (173-72)e^(6k)
137 = 72 + 101e^6k
101e^6k = 65
e^6k = 65/101
6k = ln(65/101)
k = ln(65/101)/6 = -.073455541.. (put in calculator's memory)
so when T(t) = 100
100 = 72 + 101e^((-.073...)(t))
e^((-.073..)t) = 28/101
t = ln(28/101)/-.073..) = appr 17.5 minutes
check my arithmetic
A cup of coffee at 173 degrees is poured into a mug and left in a room at 72 degrees. After 6 minutes, the coffee is 137 degrees.
Assuming that Newton's Law of Cooling applies:
After how many minutes will the coffee be 100 degrees?
2 answers
saac's law says,
T(t) = Ta + (Ts - Ta)e^kt
where Ta is the room temp
Ts is the temp of the object
and t is the time
T(6) = 72 + (173-72)e^(6k)
137 = 72 + 101e^6k
101e^6k = 65
e^6k = 65/101
6k = ln(65/101)
k = ln(65/101)/6 = -.073455541.. (put in calculator's memory)
so when T(t) = 100
100 = 72 + 101e^((-.073...)(t))
e^((-.073..)t) = 28/101
t = ln(28/101)/-.073..) = appr 17.5 minutes
check my arithmetic
T(t) = Ta + (Ts - Ta)e^kt
where Ta is the room temp
Ts is the temp of the object
and t is the time
T(6) = 72 + (173-72)e^(6k)
137 = 72 + 101e^6k
101e^6k = 65
e^6k = 65/101
6k = ln(65/101)
k = ln(65/101)/6 = -.073455541.. (put in calculator's memory)
so when T(t) = 100
100 = 72 + 101e^((-.073...)(t))
e^((-.073..)t) = 28/101
t = ln(28/101)/-.073..) = appr 17.5 minutes
check my arithmetic