**– 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:

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

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

\[Q\ =\ 189200\ J\]

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