Question

When a hot object is placed in a water bath whose temperature is 25◦C, it cools from 100 to 50◦C in 150 s. In another bath, the same cooling occurs in 120 s. Find the temperature of the second bath.

Answer 1

Apparently, you use Newton's Law of Cooling, which goes like this:

Let y(t) be the temperature of a hot object that is cooling off in an environment where the ambient temperature is T0. Newton assumed that the rate of cooling is proportional to the temperature difference y − T0. We express this hypothesis in a precise way by the differential equation
y′ = −k(y − T0) (T0 = ambient temperature)
The constant k, in units of (time)^−1, is called the cooling constant and depends on the
physical properties of the object.

Answer 2

dy/dt = -k(y - T0)
dy/(y - T0) = -kdt

Integrate both sides:

ln(y – T0) = -kt + C ----- (1)

For both baths, at time t = 0, y = 100 degrees. Substitute and solve for C

ln(100 – T0) = 0 + C;

Take C to the LHS in equation (1). Make use of ln(a) - ln(b) = ln(a/b) to simplify.

Substitute the parameters of the first bath and find the constant k.

Substitute the parameters of the second bath and find T0.