Chapter 14, Problem 25P

To determine

If a person can give off 185 kCal of heat in 25 minutes by evaporation of water (at 20⁰ C)from the skin, then how much water gets lost.

Answer to Problem 25P

Solution:

0.2984 kg water is lost by the skin.

Explanation of Solution

Given:

$\begin{array}{l}Q=\text{185 kCal}\\ Q=\text{185}\times \text{1}{0}^{\text{6}}\text{Cal}\end{array}$

Formula used:

The heat energy absorbed or released to raise the temperature by one degree is given by;

$Q=mc\Delta T$

Here,

m = mass of the substance.

C = specific heat.

◻ T= Temperature difference.

Calculation:

Evaporation of sweating starts at 100⁰C.

$\begin{array}{l}\Delta T=\text{1}{00}^0\text{C}-\text{2}{0}^0\text{C}\\ \Delta T=\text{8}{0}^0\text{C} \end{array}$

Now, the heat needed to convert water (100⁰C)to steam (at 100⁰C)is given by;

$\begin{array}{l}{{\displaystyle Q}}_2=mL\\ {{\displaystyle Q}}_2=m(2260\text{ kJ/kg)}\end{array}$

So, total heat required is given by;

$\begin{array}{l}{{\displaystyle Q=Q}}_1+{{\displaystyle Q}}_2\\ Q=(334.88\text{ kJ/kg +2260 kJ/kg)}\times m\end{array}$

$Q=\text{(2594}\text{.88}\text{}\text{kJ/kg)}\times m$ …(1)

$\begin{array}{l}\text{As }Q=\text{185 kCal}\\ Q=\text{185}\times \text{1}{0}^{\text{3}}\text{cal }\times \text{4}.\text{186 J}/\text{cal}\\ Q=\text{774}.\text{41}\times \text{1}{0}^{\text{3}} \text{J} \dots \left(\text{2}\right)\end{array}$

Equating (1)and (2);

Heat needed to raise the temperature at this level is given by;

$\begin{array}{l}\text{(2594}\text{.88}\text{}\text{kJ/kg)}\times m\text{=774}\text{.41 kJ}\\ m\text{=0}\text{.2984 kg}\\ \end{array}$