An energetic athlete can use up all the energy from a diet of $\mathbf{4000}\mathbf{}\mathbf{Cal}\mathbf{/}\mathbf{day}\mathbf{.}$. If he were to use up this energy at a steady rate, what is the ratio of the rate of energy use compared to that of a$\mathbf{100}\mathbf{}\mathbf{\text{W}}$bulb? (The power of$\mathbf{100}\mathbf{}\mathbf{\text{W}}$is the rate at which the bulb converts electrical energy to heat and the energy of visible light.)

1. The rate of energy used up by the athlete $=4000\text{Cal/day}$
2. The rate of conversion of energy of a bulb$=100\text{W}$

An incandescent bulb converts electrical energy partially to heat and partially to light. An athlete gains energy through diet and consumes it later. The rate of energy consumed by the athlete can be compared to the rate of energy conversion by a bulb.

Formula:

1. P= rate of conversion of energy
2. Dietary calorie,$1\text{Cal}=1000\text{cal}$
3. $1\text{cal}=4.19\text{J}$
   

The rate of energy gain by the athlete from a diet is$4000\text{Cal/day}$

Using equation (ii), the rate of energy used up the athlete is given by

$\begin{array}{c}4000\frac{\text{Cal}}{\text{day}}=4000\frac{\text{Cal}}{\text{day}}\times \frac{\text{1000cal}}{\text{1Cal}}\\ =4\times {10}^6\text{cal/day}\end{array}$

So, from the value of equation (iii), the rate of energy gained by the athlete in terms of J/s or Wis given by

Now the ratio of energy conversion rate by athlete to that of bulb is given by

$\begin{array}{c}{P}_a=\frac{{\text{4×10}}^{\text{6}}\text{cal}}{\text{day}}\times \frac{\text{1day}}{\text{24×3600s}}\times \frac{\text{4.19J}}{\text{1cal}}\\ =\frac{4\times {10}^6\times 4.19}{24\times 3600}\text{J/s}\\ =194\text{J/s}\\ =194\text{W}\end{array}$

Hence, the required ratio is.$1.94$