The temperature of a body falls from 30 deg C to 20 deg C in 5 minutes. The air temperature is 13...

Question:

The temperature of a body falls from 30° C to 20° C in 5 minutes. The air temperature is 13° C. Find its temperature after another 5 minutes.

Newton's Cooling Law

When materials are placed in an environment with a constant temperature, the temperature of the material will change to achieve thermal equilibrium in the system. The change in the materials temperature is given by an exponential function known as Newton's Cooling Law equation.

Newton's cooling law equation is given by

{eq}T(t) = T_s + (T_0 - T_s) e^{-kt} {/eq}

where T(t) is the temperature at time t, {eq}T_s {/eq} is the temperature of the surroundings, {eq}T_0 {/eq} is the initial temperature and k is the decay constant.

For this problem, we have the following given:

{eq}\begin{align*} T_ s &= 13^{\circ} C,\\ T_0 &= 30^{\circ} C \\ T(5) &= 20^{\circ} C \end{align*} {/eq}

From the given we need to calculate k to write an expression for the Newton's cooling equation for this specific problem.

To solve for k, we plug all the given into the general equation.

{eq}\displaystyle \begin{align*} T(t) &= T_s + (T_0 - T_s) e^{-kt}\\ T(5) &= 13^{\circ} C + (30^{\circ} C - 13^{\circ} C)e^{-(5)k} \\ 20^{\circ} C &= 13^{\circ} C + (17^{\circ} C)e^{-5k} \\ 20^{\circ} C - 13^{\circ} C &= (17^{\circ} C)e^{-5k} \\ \frac{20^{\circ} C - 13^{\circ} C }{ (17^{\circ} C)} &= e^{-5k} \\ e^{-5k} &= \frac{7}{17} \\ \end{align*} {/eq}

Taking the natural logarithm of both sides:

{eq}\displaystyle \begin{align*} \ln \bigg( e^{-5k} &= \frac{7}{17} \bigg) \\ -5k&= \ln \bigg( \frac{7}{17} \bigg) \\ k&= -\frac{1}{5} \ln \bigg( \frac{7}{17} \bigg) \\ \end{align*} {/eq}

Therefore, the explicit Newton's cooling equation for the problem is {eq}\displaystyle \boxed{ T(t) = 13^{\circ} C + (30^{\circ} C - 13^{\circ} C) e^{\frac{1}{5} \ln \bigg( \frac{7}{17} \bigg)t}} {/eq}.

Now if we want to know the temperature of the body after 5 more minutes, we solve the equation for t = 10.

{eq}\displaystyle \begin{align*} T(t) &= 13^{\circ} C + (30^{\circ} C - 13^{\circ} C) e^{\frac{1}{5} \ln \bigg( \frac{7}{17} \bigg)t},\ t = 10\\ T(10) &= 13^{\circ} C + (30^{\circ} C - 13^{\circ} C) e^{\frac{1}{5} \ln \bigg( \frac{7}{17} \bigg)(10)} \\ &= 13^{\circ} C + (17^{\circ} C) e^{2 \ln \bigg( \frac{7}{17} \bigg)} \\ T(10) &= \boxed{15.88^{\circ}\ C} \end{align*} {/eq}

The temperature of the body after five more minutes is {eq}\boxed{15.88^{\circ}\ C} {/eq}.