Answer to Question #101814 in Calculus for Joe

Question

In a water bath with an area of ​​A = 10.0 m^2 and depth h = 1.5 m, temperature is initially T0 = 4 ° C throughout. The water is heated so that finally the water the temperature follows the equation T (z) = T0 + 15e^+0,5z in the depth direction[in ° C]. The temperature coefficient of water volume is y = 500 x 10^-6 1/°C. How much has the pool water volume increased during heating?

Answer 1

The initial volume of the pool water equals

$$V_0=Ah=10\cdot 1.5=15m^3.$$

The final volume could be defined from

$$V_0=y(T-T_0)V,\\ V=\frac{V_0}{y(T-T_0)}.$$

Thus, an elementary volume of water of height $dz$ is given by

$$dV=\frac{Adz}{y(T(z)-T_0)},\\ dV=\frac{Adz}{y\cdot 15e^{+0.5z}}.$$

Integrate

$$V=\int^h_0\frac{Adz}{y\cdot 15e^{0.5z}},\\ V=\frac{A}{15y}\int^h_0e^{-0.5z}dz,\\ V=\frac{A}{15y}\cdot\left.\frac{e^{-0.5z}}{-0.5}\right\vert ^h_0.$$

Substitute

$$V=\frac{10}{15\cdot 500\cdot 10^{-6}}\cdot\frac{e^{-0.5\cdot 1.5}-1}{-0.5}=1407m^3$$

Thus, the pool water volume increased on $V-V_0=1407-15=\bm{1392 m^3}.$

The result is unrealistic due to unrealistic dependency of the water temperature on the depth $T (z) = T_0 + 15e^{+0,5z}$ .