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For how long a time t could a student jog before irreversible body damage occurs?

– Thermal energy is generated at a rate of $1200W$ when a student weighing $70-kg$ is running.

– This thermal energy must be dissipated from the body through perspiration or other processes to maintain the body temperature of the runner at a constant $37\ ^{ \circ }C$. In the event of failure of any such mechanism, the thermal energy would not be dissipated from the student’s body. In such a scenario, calculate the total time the student can run before his body faces irreversible damage.

– (If the body temperature rises above $44\ ^{ \circ }C$, it caused irreversible damage to the protein structure in the body. A standard human body has slightly lower specific heat than that of water i.e. $3480\ \dfrac{J}{Kg. K}$. The presence of fat, proteins, and mineral in the human body causes the difference in specific heat as these components have specific heats of lower value.)

The aim of this question is to find the time a student can run continuously before causing his body to overheat and result in irreversible damage.

The basic concept behind this article is Heat Capacity and Specific Heat.

Heat Capacity $Q$ is defined as the quantity of heat that is required to cause a temperature change of the given quantity of a substance by $1^{ \circ }C$. It can either be heat discharged or heat gained by the substance. It is calculated as follows:

\[Q=mC∆T\]

Where:

$Q=$ Heat Capacity (Heat discharged or gained by the body)

$m=$ Mass of the Substance

$C=$ Specific Heat of the Substance

$∆T=$ Temperature Difference $=T_{Final}-T_{Initial}$

Expert Answer

Given that:

Initial Temperature $T_1=37^{ \circ }C=37+273=310K$

Raised Temperature $T_2=44^{ \circ }C=44+273=317K$

Mass of Student $m=70Kg$

Rate of Thermal Energy $P=1200W$

Specific Heat of Human Body $C=3480\frac{J}{Kg. K}$

The heat generated by the human body as a result of running is calculated as follows:

\[Q=mC∆T=mC(T2-T1)\]

\[Q=70Kg\times(3480\frac{J}{Kg.K})(317K-310K)\]

\[Q\ =\ 1705200\ \ J\]

\[Q\ =\ 1.705\times{10}^6J\]

The Rate of Thermal Energy Generation is calculated as follows:

\[P\ =\ \frac{Q}{t}\]

\[t\ =\ \frac{Q}{P}\]

\[t\ =\ \frac{1.705\times{10}^6\ J}{1200\ W}\]

As we know:

\[1\ W\ =\ 1\ \frac{J}{s}\]

So:

\[t\ =\ \frac{1.705\times{10}^6\ J}{1200\ \frac{J}{s}}\]

\[t\ =\ 1421\ s\]

\[t\ =\ \frac{1421}{60}\ min\]

\[t\ =\ 23.68\ min\]

Numerical Result

The total time the student can run before his body faces irreversible damage is:

\[t\ =\ 23.68\ min\]

Example

A cube having a mass of $400g$ and specific heat of $8600\ \frac{J}{Kg. K}$ is initially at $25 ^{ \circ }C$. Calculate the amount of heat that is required to raise its temperature to $80 ^{ \circ }C$.

Solution

Given that:

Mass of the cube $m\ =\ 400\ g\ =\ 0.4\ Kg$

The Specific Heat of Cube $C\ =\ 8600\ \frac{J}{Kg. K}$

Initial Temperature $T_1\ =\ 25 ^{ \circ }C\ =\ 25+273\ =\ 298\ K$

Raised Temperature $T_2\ =\ 80 ^{ \circ }C\ =\ 80+273\ =\ 353\ K$

The amount of heat that is required to raise its temperature is calculated as per the following formula:

\[Q\ =\ mC∆T = mC(T2-T1)\]

Substituting the values in the above equation:

\[Q\ =\ (0.4\ Kg)(8600\ \frac{J}{Kg.K})(353\ K-298\ K)\]

\[Q\ =\ (0.4\ Kg)(8600\ \frac{J}{Kg.K})(55\ K)\]

\[Q\ =\ 189200\ J\]

\[Q\ =\ 1.892\times{10}^5\ J\]

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