dT/dt=0.1(34-T)
dT/(34-T) = 0.1dt
Integrate both sides
-ln(34-T)=0.1t+c
34-T=Ce^(-0.1t)
T=34-Ce^(-0.1t)
Initial conditions:
t=0, T=4 => 4=34-C => C=30
so
T=34-30e^(0.1t)
A) Substitute t=15 to solve for T.
B) Substitute T=30 to solve for t.
on a day when the air temperature is 34 degrees celsius, a can of soft drink is taken from the fridge at a temperature of 4 degrees celsius at 2pm. Thereafter the temperature of the can satisfies the differential equation; dT/dt=0.1(34-T), where t is the number of minutes after 2pm. What is the temperature of the can at 2:15pm? At what time does the temperature reach 30 degrees?
3 answers
wow thank you so much
You're welcome!
2 answers