Hi, i can help with this.
first thing is to break down the information and work one at a time. so first we're given the following "initial temperature of 95°C and is placed on a sink in the kitchen where the temperature is a constant 25°C" i.e at t=0, T=95 with Ts=25.
subbing it into the formula gives:
dT/dt = -k(T-25)
Next is to flip the derivative to make t the subject when integrating.
dt/dT = -1/k(T-25)
integrating both sides gives,
t=-1/k * ln(T-25)+c where T>25.
now we can sub t=0 and T=95, which is,
0=-1/k * ln(70)+c
i.e. c=1/k * ln(70)
subbing it into the original equation gives,
t=-1/k * ln(T-25)+1/k * ln(70)
t= 1/k * ln(70/(T-25))
now we can solve for k where the new information tells us "it takes 10 minutes to reach a temperature of 60°C" which is when t=10, then T=60. subbing these in gives,
10=1/k * ln(70/(60-25))
10=1/k * ln(2)
k= ln(2)/10
finally we can sub this into our final equation of t= 1/k * ln(70/(T-25)),
t=10/ln(2)ln(70/(T-25))
but the questions asks to express T in terms of t. so just rearrange it.
(t/10)ln(2)=ln(70/(T-25))
ln(2t/10)=ln(70/T-25))
2t/10=70/(T-25)
T=70/2t/10+25 or T=70*2-t/10+25.
this can be checked where t=10 which gives T=60.
A good way to remember or understand is if u have the Cambridge textbook, go to chapter 11D. Applications of differential equations. example 15.
you can message me on discord if u need anymore help 🙂 (user is jomjom7461)
