# High altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate energy absorbed from your body if you: (more information see details)?

## a.) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius (ANSWER: 5.210^5J)? b.) melt 1.0kg of -15 degrees Celsius snow using a stove and drink the resulting 1.0kg of water at 2 degrees Celsius, which your body has to warm to 37 degrees Celsius (ANSWER: 1.510^5)?

Feb 11, 2018

(a)

1. Heat required to change $1.0 \setminus k g$ ice at $- {15}^{\circ} C$ to Ice at ${0}^{\circ} C$ $= m {s}_{i c e} \Delta T = 1 \times \left(2.108 \times {10}^{3}\right) \times 15 = 3.162 \times {10}^{4} \setminus J$
2. Heat required to change from state of $1.0 \setminus k g$ ice to water$= m {L}_{i c e} = 1 \times 3.34 \times {10}^{5} = 3.34 \times {10}^{5} \setminus J$
3. Heat required to change $1.0 \setminus k g$ water at ${0}^{\circ} C$ to water at ${37}^{\circ} C$ $= m {s}_{w a t e r} \Delta T = 1 \times \left(4.187 \times {10}^{3}\right) \times 37 = 1.549 \times {10}^{5} \setminus J$

Total heat required $= 3.162 \times {10}^{4} + 3.34 \times {10}^{5} + 1.549 \times {10}^{5}$
$= 5.21 \times {10}^{5} \setminus J$

(b)

Heat required to warm $1.0 \setminus k g$ water at ${2}^{\circ} C$ to water at ${37}^{\circ} C$ $= m {s}_{w a t e r} \Delta T = 1 \times \left(4.187 \times {10}^{3}\right) \times 35 = 1.465 \times {10}^{5} \setminus J$