#### Answer

(a) The average power is $94.0~Watts$
(b) The chemical energy that is used in this hike is $2.03\times 10^6~J$
(c) A person would use 485 kilocalories of food energy.

#### Work Step by Step

(a) We can find the increase in gravitational potential energy:
$mgh = (70~kg)(9.80~m/s^2)(740~m) = 507,640~J$
We can find the average power output:
$Power = \frac{Energy}{Time} = \frac{507,640~J}{(1.5)(3600~s)} = 94.0~Watts$
The average power is $94.0~Watts$
(b) The mechanical energy output is 25% of the chemical energy that is used in this hike. We can find the chemical energy $E_c$ that is used:
$0.25E_c = 507,640~J$
$E_c = \frac{507,640~J}{0.25}$
$E_c = 2.03\times 10^6~J$
The chemical energy that is used in this hike is $2.03\times 10^6~J$
(c) We can convert the chemical energy to units of kilocalories:
$E_c = 2.03\times 10^6~J \times \frac{1~kcal}{4184~J} = 485~kcal$
A person would use 485 kilocalories of food energy.