We can solve this problem using separation of variables.
Then apply the initial conditions
EXPLANATION
We were given the first order differential equation

We now separate the time and the temperature variables as follows,

Integrating both sides of the differential equation, we obtain;

This natural logarithmic equation can be rewritten as;

Applying the laws of exponents, we obtain,


We were given the initial conditions,

Let us apply this condition to obtain;


Now our equation, becomes

or

When we substitute a=20,
we obtain,

b) We were also given that,

Let us apply this condition again to find k.

This implied



We take logarithm to base e of both sides,

This implies that,


After 15 minutes, the temperature will be,



After 15 minutes, the temperature is approximately 20°C