What mass of butter, which has a usable energy content of$\mathbf{\text{6.0Cal/g}}\mathbf{\left(}\mathbf{\text{=6000cal/g}}\mathbf{\right)}\mathbf{\text{,}}$would be equivalent to the change in gravitational potential energy of a$\mathbf{\text{73.0kg}}$man who ascends from sea level to the top of Mt. Everest, at elevation$\mathbf{\text{8.84km}}$? Assume that the average g for the ascent is${\mathbf{\text{9.80m/s}}}^{\mathbf{\text{2}}}$.

1. The heat of transformation of butter is$\mathrm{L}=6000\text{\hspace{0.17em}cal/g}$
2. The mass of the man is$\mathrm{m}=73\text{\hspace{0.17em}kgor}73000\text{\hspace{0.17em}g}$
3. The height from which the man climbs Mt. Everest is$\mathrm{h}=8.84\text{kmor}8.84\times {10}^3\text{m}$
4. The acceleration due to gravity for the ascent is$\mathrm{g}=9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$

The energy that an object has due to its location in a gravitational field is known as gravitational potential energy. The amount of energy released or absorbed during a change in state is known as a substance's latent heat. We can use the concept of gravitational potential energy. When the man climbs to the top of Mt. Everest, he gets gravitational potential energy. His energy is converted to calories. By using the concept of the heat of transformation, we can find the amount of butter of usable energy.

Formulae:

The potential energy of the body, $U=mgh$ …(i)

The heat energy released or absorbed by the body,$\mathrm{Q}=\mathrm{Lm}$ …(ii)

Where, m = mass

L = specific latent heat

The mass of the butter which has usable energy content:

The energy used by the man to climb to the top of Mt. Everest using equation (i) is given as:

$\begin{array}{c}\mathrm{U}=73.0\text{kg}\times 9.8{\text{m/s}}^{\text{2}}\times 8.84\times {10}^3\text{m}\\ =6.32\times {10}^6\text{J}\end{array}$

Here, it is given that the gravitational potential energy is equal to the heat released, so:

$\begin{array}{c}\mathrm{Q}=\mathrm{U}\\ =6.32\times {10}^6\text{J}\\ =\frac{(6.32\times {10}^6\text{J})\times (1.00\text{cal})}{4.186\text{J}}\text{(}\because 1\text{cal}=4.186\text{J)}\end{array}$

Using equation (ii) and the given values, we get the mass of the butter as given:

$\begin{array}{c}\mathrm{m}=\frac{\frac{(6.32\times {10}^6\mathrm{J})\times (1.00\text{cal})}{4.186\text{J}}}{6000\text{cal/g}}\\ =250\text{g}\end{array}$

Hence, the value of the mass of the butter is$250\text{g}$