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.
[tex]ms_{ice}[/tex]ΔT = (1.0)(2.100×10³)(15) = 3.150×[tex]10^{4}[/tex]J
Heat required to convert 1.0 kg ice to water.
[tex]mL_{ice} = 1[/tex]×[tex]3.33[/tex]×[tex]10^{5}[/tex] = 3.33×[tex]10^{5}[/tex]J
Heat required to convert 1.0 kg water at 0°C to 37°C.
[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×37 = 1.548×[tex]10^{5}[/tex] J
Total heat required = 3.150×[tex]10^{4}[/tex] + 3.33×[tex]10^{5}[/tex] + 1.548×[tex]10^{5}[/tex]
= 5.19×[tex]10^{5}[/tex] J
b)
Heat required to warm 1.0 kg water at 2°C to water at 37°C.
[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×35
= 1.465×[tex]10^{5}[/tex]J
