Newton's law of cooling states that ‘the rate of cooling of a body is proportional to the difference between the temperature of a body and the temperature of its surroundings.’ If θ is the difference between the temperature of a body and the temperature of its surroundings then dθ/dt = -kθ - Leaving Cert Applied Maths - Question 10 - 2012

Continuing from the relationship established, we manipulate it further. We substitute t = 25 minutes into the equation:

$\ln \theta = -kt + C$

We already have k and need to find C using the initial conditions:

$\theta(0) = 80$.

This results in:

$C = \ln 80$.

Now substituting k and C back into the equation:

$\ln \theta(25) = -25k + \ln 80$

Calculating the terms provides:

$\theta(25) = e^{-25k} \cdot 80$

Substituting in the numerical value of k:

$\theta(25) \approx e^{-1.0125} \cdot 80$

This gives us:

$\theta(25) \approx 41.78° C$

From the context provided, we start with the equation of motion through the resistive gel:

$\frac{dv}{dt} = -kv^2$

To separate variables and integrate:

$\int_{1000}^{10} \frac{1}{v^2} dv = -k \int_{0}^{t} dt$

Integrating gives:

$\left[-\frac{1}{v}\right]_{1000}^{10} = -kt$

Plugging in values:

$-\left(\frac{1}{10} - \frac{1}{1000}\right) = -kt$

Consequently, you find:

$k = \frac{99}{10}$.

Here we have another differential equation:

$\frac{dr}{dt} = \frac{1}{9900t + 1}$

Integrating gives:

$\int dr = \int \frac{1}{9900t + 1} dt$

The solution to this integral leads us to:

$r = \frac{1}{9900} ln(9900t + 1) + C$

Using the limits provided and combining knowledge of motion, we find:

Inserting values respectively, we conclude that:

The length of the gel block formulates to: $L = \frac{10}{99} \ln 100.$