Question

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg?C?, the latent heat of fusion is 333 kJ/kg, the specific heat of water is4186 J/kg?C?.
a) Calculate the energy absorbed from a climber's body if he eats 0.90
kg of -15?C snow which his body warms to body temperature of 37?C.
b) Calculate the energy absorbed from a climber's body if he melts 0.90
kg of -15?C snow using a stove and drink the resulting 0.90kg of water at 2?C, which his body has to warm to 37?C.

Answer 1

Answer:

a) $Q=467443.8\ J$

b) $Q_m=299700\ J$

c) $Q_2=131859\ J$

Explanation:

Given:

• specific heat of ice, $c_i=2100\ J.kg^{-1}.^{\circ}C^{-1}$

• latent heat of fusion of ice, $L=333000\ J.kg^{-1}$

• specific heat of water, $c_w=4186\ J.kg^{-1}.^{\circ}C^{-1}$

(a)

• mass of snow, $m_s=0.9\ kg$

• initial temperature of snow, $T_{is}=-15^{\circ}C$

• Final temperature of the consumed mass, $T_f=37^{\circ}C$

Now the energy absorbed from the body after eating this snow:

$Q=m_s.c_i.\Delta T_i+m_s.L+m_s.c_w.\Delta T_w$

$Q=0.9\times 2100\times (0-(-15))+0.9\times 333000+0.9\times 4186\times (37-0)$

$Q=467443.8\ J$

(b)

Energy absorbed from the body in melting the ice is the total latent heat:

$Q_m=m_s.L$

$Q_m=0.9\times 333000$

$Q_m=299700\ J$

(c)

• initial temperature of water, $T_{iw}=2^{\circ}C$

• final temperature of water, $T_{iw}=37^{\circ}C$

Now, the amount of energy invested by body for the water at this condition:

$Q_2=m_s.c_w.\Delta T_2$

$Q_2=0.9\times 4186\times (37-2)$

$Q_2=131859\ J$