# Newton's law of cooling: changing temperature of environment

A metal ball having temperature of $80^\circ C$ is placed into $m$ grams of water at $0^\circ C$. After ten minutes, it was found that the temperature of ball and water are $60^\circ C$ and $20^\circ C$ respectively. After that the ball (at $60^\circ C$) is placed into another $m$ grams of water at $0^\circ C$. Find the temperature after $10$ minutes.

I tried approaching this by Modification of Newton's Law of Cooling with two initial conditions, I was able to eliminate the constants of integration I am not being able to eliminate the heat transfer coefficient with latter two initial conditions, I am not being able to (Mathematica returns no solution). How do I determinate the temperature after $10$ minutes? Is it not possible to eliminate coefficient of heat transfer with initial conditions?

I don't feel like making this overly complicated. The heat flow is proportional to temperature difference so the the temperature difference decays exponentially. Then if it halves in the first ten minutes, it will halve in the second ten minutes. From the first part we know the sphere and water have the same heat capacity so their temperatures changes are equal. So the sphere winds up and 45$^\circ$C and the water at 15$^\circ$C.

# Using differential equations

If it is a requirement that you have to use differential equations to solve the problem, then I will redo the first part in terms of differential equations. It's the same logic but I will justify each statement with a differential equation.

Let $T_b(t)$ and $T_w(t)$ be the temperature of the ball and water respectively. Let $t_0$ be the time the ball is put in the water and $t_1$ be ten minutes later. Let $t_2$ be the time the ball is put into the other, and $t_3$ be ten minutes after that.

Then we have $$\dot{T}_b = C_{b1}(T_w - T_b)$$ $$\dot{T}_w = C_{w1}(T_b - T_w)$$ $$T_b(t_0) = 80^\circ C$$ $$T_w(t_0) = 0^\circ C$$ $$T_b(t_1) = 60^\circ C$$ $$T_w(t_1) = 20^\circ C$$

and

$$\dot{T}_b = C_{b2}(T_w - T_b)$$ $$\dot{T}_w = C_{w2}(T_b - T_w)$$ $$T_b(t_2) = 60^\circ C$$ $$T_w(t_2) = 0^\circ C$$.

The first observation to make is that since the two volumes of water have the same mass, $C_{b1} = C_{b2} = C_b$ and $C_{w1} = C_{w2} = C_w$.

The first statement I made is that the temperature difference decays exponentially. This is apparent because taking the difference of the equation for $\dot{T}_b$ and $\dot{T}_w$, we get $$\frac{d}{dt} (T_b - T_w ) = -(C_b + C_w)(T_b - T_w).$$

Thus for the first part $T_b - T_w = (T_b(t_0) - T_w(t_0))\exp(-(C_b + C_w)(t-t_0))$ and so
$(T_b(t_0) - T_w(t_0))\exp(-(C_b + C_w)(t_1-t_0)) = T_b(t_1) - T_w(t_1) = \frac{1}{2}(T_b(t_0) - T_w(t_0))$. It must be that the temperature difference halves in ten minutes: $\exp(-(C_b + C_w)(t_1-t_0)) = \frac{1}{2} = \exp(-(C_b + C_w)(t_3-t_2))$.

Thus $T_b(t_3) - T_w(t_3) =(T_b(t_2) - T_w(t_2))\exp(-(C_b + C_w)(t_3-t_2)) = \frac{1}{2}(T_b(t_2) - T_w(t_2))$. This says the final temperature difference after the second ten minutes must be $30^\circ C$.

The next claim I made is that the water and the ball have the same heat capacity. This is evident since the sum of the temperatures doesn't change. This time we add equations instead of subtracting:

$$\frac{d}{dt} (T_b + T_w ) = -(C_b - C_w)(T_b - T_w).$$

Thus $T_b + T_w = (T_b(t_0) + T_w(t_0))\exp(-(C_b - C_w)(t-t_0))$ and so
$(T_b(t_0) + T_w(t_0))\exp(-(C_b - C_w)(t_1-t_0)) = T_b(t_1) + T_w(t_1)$. Since $T_b(t_0) + T_w(t_0) = T_b(t_1) + T_w(t_1)$, it must be that $\exp(-(C_b - C_w)(t_1-t_0))=1$ and so $C_b = C_w$, and the sum of the temperatures remains constant.

Thus we have $T_b(t_3)-T_w(t_3) = 30^\circ C$ and $T_b(t_3)+T_w(t_3) = 60^\circ C$, so $T_b(t_3) = 45^\circ C$ and $T_w(t_3) = 15^\circ C$.

• Actually this is differential equation h/w question in mathematical physics. If I write above logic, I would get zero marks. How do I solve that first order system mathematically? – hasExams Jun 2 '13 at 0:27