# Captain Schultz, the star high diver of the circus, insists that his bath always be at exactly 37...

## Question:

Captain Schultz, the star high diver of the circus, insists that his bath always be at exactly 37 C, body temperature, and measures the temperature of the bath with a precision thermometer before relaxing in it at the end of his day?s work. One terrible evening not so long ago the captain found his bath to be only 36.8 C. His noble countenance glowed with fury as he ordered his trembling valet to move his portable 80 kg tub from his private dressing tent to the foot of the high-dive ladder. For a moment he paused, deep in thought, and then, ignoring the murmurs of the gathering crowd, he mounted the vertical ladder, counting the rungs as he climbed. He stopped at what was obviously a critical height. Disdainfully he let his dressing gown fall to the spellbound throng far below, and launched himself into the air, landing in his tub with such infinite precision that nary a drop of water was lost from the tiny splash. A smile lighted his features as he relaxed in his 37 C tub, while his valet quickly wheeled him back to his tent. Captain Schultz is 170 cm tall, has blond hair, blue eyes, weighs 70 kg. How high did he climb? Assume that water has a constant heat capacity and estimate changes in internal energy using {eq}\Delta {/eq}U = mCv{eq}\Delta {/eq}T.

## Conservation of energy

Conservation of energy principle states that the energy of state 1 is equal to the energy of state 2. only the form changes, but the total energy remains at the start point and the end point constant.

Given Data

• The mass of the captain Schultz is: {eq}m = 70\;{\rm{kg}} {/eq}.
• The mass of the tub is: {eq}M = 80\;{\rm{kg}} {/eq}.
• The initial temperature is: {eq}{T_1} = \left( {37 + 273} \right)\;{\rm{K = 310}}\;{\rm{K}} {/eq}.
• The final temperature is: {eq}{T_2} = \left( {36.8 + 273} \right)\;{\rm{K}} = 309.8\;{\rm{K}} {/eq}.

The expression for the conservation of energy is,

{eq}mgh = M{c_v}\left( {{T_2} - {T_1}} \right) {/eq}

Substitute the given values.

{eq}\begin{align*} \left( {70\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)h &= \left( {80\;{\rm{kg}}} \right)\left( {4.180\;{\rm{kJ/kg}}{\rm{.K}}} \right)\left( {\dfrac{{1000\;{\rm{J}}}}{{1\;{\rm{kJ}}}}} \right)\left( {310 - 309.8} \right)\;{\rm{K}}\\ {\rm{h}} &= \dfrac{{66880}}{{686.7}}\\ &= 97.3933\;{\rm{m}} \end{align*} {/eq}

Thus he climb up to 97.3933 m.