Question: (I) How much energy would the person of Example 15–7 transform if instead of working 11.0 h she took a noontime break and ran at \({\bf{15}}\;{{{\bf{km}}} \mathord{\left/{\vphantom {{{\bf{km}}} {\bf{h}}}} \right.} {\bf{h}}}\) for 1.0 h?

The energy transforming process that may occur in the body is known as the metabolism process. There are two kinds of metabolism processes: (i) Catabolism and (ii) Anabolism.

From table 15-2:

The metabolic rate for sitting upright is\(115\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}\).

The metabolic rate for running \(15\;{{{\rm{km}}} \mathord{\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}\) is \(1150\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}\).

In the example problem, the result indicate the total energy transformed in 24 hours is\(1.15 \times {10^7}\;{\rm{J}}\).

Instead of working 11.0 h, she took noon break and ran for 1.0 hour. Therefore, subtract energy for one hour of sitting upright and add energy for 1 hour of running at\(15\;{{{\rm{km}}} \mathord{\left/{\vphantom {{{\rm{km}}} {\rm{h}}}} \right.} {\rm{h}}}\).

Therefore, the total energy transformed by the person can be calculated as:

\(\begin{aligned}{c}E &= \left( {1.15 \times {{10}^7}\;{\rm{J}}} \right) + \left( { - \left( {115\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}} \right) + \left( {1150\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {\rm{s}}}} \right.} {\rm{s}}}} \right)} \right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)\\E &= 1.52 \times {10^7}\;{\rm{J}}\end{aligned}\)

Thus, the total energy transformed in this process is \(1.52 \times {10^7}\;{\rm{J}}\).