Question

A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at$\mathbf{37}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathit{C}$. Suppose that the water were to cool down spontaneously to form ice at$\mathbf{0}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}\mathit{C}\mathbf{,}$, and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20, this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

Short answer

a) The bather will reach a height of 260 km.

Step-by-step solution

Step 1:Given Data

Mass of person: $70\mathrm{kg}$

Dimension of bathtub: $1.90\hspace{0.33em}\mathrm{m}\times 0.8\mathrm{m}$

Height to which water is filled: $0.27\mathrm{m}$

Initial temperature of water: $37.0^{°}\mathrm{C}$

Final temperature of water: $0^{°}\mathrm{C}$

Concept of Latent Heat.

Latent Heat is associated with the amount of energy either absorbed or released during phase change of a substance, i.e. from liquid to solid, from solid to liquid, from liquid to gas, from gas to liquid or from gas to solid and solid to gas.

Determination of the height to which the bather would go.

The equation will correspond to the heat energy that flows into the water will be equivalent to the resulting gravitational potential energy.

$L_f{\rho }_w{V}_w+c_w{\rho }_w{V}_w\Delta T=mgh$

Here, h is the maximum height reached, m is the mass of bather, g is the acceleration due to gravity of earth, $L_f$is the latent heat of fusion, $p_w$is the density of water, $V_w$is the volume of water, $C_w$is the specific heat of water and $∆T$is the change in temperature.

Rearrange for h and substitute all the values,

$$\begin{gathered} \mathrm{h}=\frac{\left(1000\mathrm{kg}/\mathrm{m}\right)\left(1.90\mathrm{m}\times 0.800\mathrm{m}\times 0.240\mathrm{m}\right)\left[33410\mathrm{J}/\mathrm{kg}+\left(4190\mathrm{J}/\mathrm{kg}\mathrm{K}\right)\left(37.0^{°}\mathrm{C}\right)\right]}{\left(70.0\mathrm{kg}\right)\left(9.80\mathrm{m}/\mathrm{s}\right)} \\ =260\mathrm{km} \end{gathered}$$

Thus the height to which the bather will go is 260 km.