Understanding Newton's Law of Cooling

The problem describes a situation governed by Newton's Law of Cooling, which states that the rate of temperature change of an object is proportional to the difference between its own temperature and the ambient temperature. Mathematically, this is expressed as:

$ \frac{dT}{dt} = -k(T(t) - T_{amb}) $

Where:

• $T(t)$ is the temperature of the object at time $t$.
• $T_{amb}$ is the ambient temperature.
• $k$ is a positive constant related to the object's properties and environment.

Applying the Law to the Heat Exchanger

From the problem statement:

• Initial temperature of the heat exchanger when cooling begins (at $t=0$): $T(0) = 90^{\circ}\text{C}$.
• Ambient temperature: $T_{amb} = 30^{\circ}\text{C}$.
• The differential equation becomes: $ \frac{dT}{dt} = -k(T(t) - 30) $.

Solving the Differential Equation

Integrating the differential equation yields the general solution:

$ T(t) = T_{amb} + C e^{-kt} $

Substituting $T_{amb} = 30^{\circ}\text{C}$:

$ T(t) = 30 + C e^{-kt} $

Determining the Constant C

Using the initial condition $T(0) = 90^{\circ}\text{C}$:

$ 90 = 30 + C e^{-k \cdot 0} $

$ 90 = 30 + C $

$ C = 60 $

The specific solution becomes:

$ T(t) = 30 + 60 e^{-kt} $

Calculating the Cooling Constant k

We are given that at $t = 30$ min, the temperature $T(30) = 70^{\circ}\text{C}$. Using this in the specific solution:

$ 70 = 30 + 60 e^{-k \cdot 30} $

$ 40 = 60 e^{-30k} $

$ e^{-30k} = \frac{40}{60} = \frac{2}{3} $

To find $k$, we can take the natural logarithm of both sides:

$ -30k = \ln\left(\frac{2}{3}\right) $

$ k = -\frac{1}{30} \ln\left(\frac{2}{3}\right) = \frac{1}{30} \ln\left(\frac{3}{2}\right) $

Predicting Temperature at t = 51.5 min

Now we need to find the temperature $T(51.5)$ using the derived formula and the calculated $k$:

$ T(51.5) = 30 + 60 e^{-k \cdot 51.5} $

Substitute the value of $k$:

$ T(51.5) = 30 + 60 e^{-\left(\frac{1}{30} \ln\left(\frac{3}{2}\right)\right) \cdot 51.5} $

$ T(51.5) = 30 + 60 e^{-\frac{51.5}{30} \ln\left(\frac{3}{2}\right)} $

Using the logarithm property $a \ln(b) = \ln(b^a)$:

$ T(51.5) = 30 + 60 e^{\ln\left(\left(\frac{3}{2}\right)^{-\frac{51.5}{30}}\right)} $

Since $e^{\ln(x)} = x$:

$ T(51.5) = 30 + 60 \left(\frac{3}{2}\right)^{-\frac{51.5}{30}} $

Calculate the value:

• $ \left(\frac{3}{2}\right)^{-\frac{51.5}{30}} = (1.5)^{-1.71666...} \approx 0.45691 $
• $ T(51.5) \approx 30 + 60 \times 0.45691 $
• $ T(51.5) \approx 30 + 27.4146 $
• $ T(51.5) \approx 57.4146^{\circ}\text{C} $

Alternatively, calculating $k$ first: $k = \frac{\ln(1.5)}{30} \approx \frac{0.405465}{30} \approx 0.0135155$.

Then, $-k \times 51.5 \approx -0.0135155 \times 51.5 \approx -0.69604$.

$e^{-0.69604} \approx 0.49860$.

$T(51.5) \approx 30 + 60 \times 0.49860 \approx 30 + 29.916 \approx 59.916^{\circ}\text{C}$.

Rounding off to the nearest integer gives 60$^{\circ}$C.