What time will its temperature reach $30^◦C$? 
Assume that the rate at which a body cools is proportional to the difference in temperature between the body and its surroundings. A body is heated to $110◦C$ and is placed in air at $10◦C$. After one hour, its temperature is $60◦C.$ At what time will its temperature reach $30◦C$?

My attempt :By the formula  $$T(t)-T(s)=(T(0)-T(s))e^{-kt}$$
$\color{orange}{\text{Given condition}}$: It takes $t=60\ minutes $ for temperature fall $\color{red}{110^\circ \ C \to 10^\circ \ C}$
Setting the corresponding values, final temperature $T(t)=10^\circ\ C$ , initial temperature $T(0)=110^\circ\ C$ & surrounding body temperature $T(s)=60^\circ\ C$  we get    $$10-60=(110-60)e^{-k(60)}$$
$$e^{-60k}=-1$$
$$-60k=\log(-1)$$
$$k=\frac{1}{-60}\log(-1)\tag 1$$
Now, for temperature fall $\color{red}{110^\circ \ C \to 30^\circ \ C}$ , setting $T(t)=30^\circ \ C$, $T(0)=110^\circ\ C$ & $T(s)=10^\circ\ C$ we get 
$$30-10=(110-10)e^{-kt}$$
$$e^{-kt}=\frac{20}{100}=\frac{1}{5}$$
$$\implies -kt=\ln\left(\frac{1}{5}\right)$$
Setting the value of $k$ from (1), we get time $t$ as follows
$$t\frac{1}{-60}\log(-1)=\log \frac{1}{5}$$
that is $$t=\log \frac{1}{5} \frac{-60}{\log(-1)}$$
Is my answer is correct or not ?
 A: The way you used your variables in the question gave the impression that you aren't familiar with them, so I would try to help you out in that regime. (Otherwise this question in it's current form is Off-topic)
The question can be easily dealt with a proper understanding of Newton's law of cooling. One of the assumptions there is that the rate of heat flow is proportional to the temperature gradient between the object and the surrounding i.e., 
$$\frac {dQ}{dt} = -k(T - T_{s})$$
Here $T$ is the temperature of the body and $T_s$ is the temperature of the surrounding. The equation in it's final form looks like the following :
$$\ln \left (\frac {T_i-T_s}{T_f-Ts} \right ) = \alpha t$$
Here $\alpha$ is constant due to the assumption that molar specific heat capacity of the substance is independent of the temperature i.e., is constant. $T_i$ is the initial temperature and $T_f$ is the final temperature. As of your question, $T_i = 110°\text C$, $T_f = 30°\text C$ and $T_s = 10°\text C$.
One important thing to note here is that when $T_f = T_s$ then the time it requires for the body to cool down approaches infinity which deviates from the reality that an object reaches to equilibrium with the surrounding in a finite time. 
