Chapter 22, Problem 110A

To determine

Increase in entropy of the universe.

To compare: The entropy increased obtained with the entropy increased due to melting of ice.

Answer to Problem 110A

Change in entropy of universe due to energy burn by person is $\Delta S=2.7\times {10}^4\text{J/K}$ and

the entropy increased by energy burn is $5.15\times {10}^4\text{J/K}$ higher than melting 20 kg of ice.

Explanation of Solution

Given:

Energy burn by person per day i.e. change in enthalpy, $\Delta H=-8.4\times {10}^6\text{J}$

Weight of ice to be melt is $W=20\text{kg}$

Formula Used:

The change in entropy is given as $\Delta S=-\frac{\Delta H}{T}$

Where, $\Delta H$ is the change in enthalpy and $T$ is the temperature.

Calculation:

As body temperature is 310 K

So, the change in entropy due to heat released by body is

  $\begin{array}{l}\Delta S=-\frac{-8.4\times {10}^6}{310}\\ =2.7\times {10}^4\text{J/K}\end{array}$

Now, for melting ice

The latent heat of melting of ice at 273 K is $3.34\times {10}^5\text{J/kg}$

So, heat required for melting 20 kg ice i.e. change in enthalpy is

  $\Delta H=20\times 3.34\times {10}^5\text{J}$

Now, change in entropy due to melting of ice is

  $\begin{array}{l}\Delta S=-\frac{20\times 3.34\times {10}^5}{273}\\ =-2.45\times {10}^4\text{J/K}\end{array}$

On comparing both entropy, the difference is

  $\begin{array}{l}2.7\times {10}^4\text{J/K}-\left(-2.45\times {10}^4\text{J/K}\right)\\ =5.15\times {10}^4\text{J/K}\end{array}$

It means the entropy increased by energy burn is $5.15\times {10}^4\text{J/K}$ higher than melting 20 kg of ice.

Conclusion:

Therefore, an increase in entropy of universe due to energy burn by person is $\Delta S=2.7\times {10}^4\text{J/K}$ and the entropy increased by energy burn is $5.15\times {10}^4\text{J/K}$ higher than melting 20 kg of ice.