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 is 4186 J/kg⋅C∘

Answer 1

Answer:

The conditions are not given in the question. Here is the complete 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 is 4186 J/kg⋅C∘

a) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius ?

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 ?

Explanation:

Let's calculate the heat required to convert ice from -15°C to 0°C and then the heat required to convert it into the water.

a)

Heat required to convert -15°C ice to 0°C.

  $ms_{ice}$ΔT = (1.0)(2.100×10³)(15) = 3.150×$10^{4}$J

Heat required to convert 1.0 kg ice to water.

    $mL_{ice} = 1$×$3.33$×$10^{5}$ = 3.33×$10^{5}$J

Heat required to convert 1.0 kg water at 0°C to 37°C.

   $ms_{water}$ΔT = 1×4.186×$10^{3}$×37 = 1.548×$10^{5}$ J

Total heat required = 3.150×$10^{4}$ + 3.33×$10^{5}$ + 1.548×$10^{5}$

                              = 5.19×$10^{5}$ J

b)

Heat required to warm 1.0 kg water at 2°C to water at 37°C.

$ms_{water}$ΔT = 1×4.186×$10^{3}$×35

                = 1.465×$10^{5}$J