Final answer:
To calculate the energy absorbed when warming 0.75 kg of snow from -15°C to body temperature, we find the energy needed to raise the temperature of the snow to 0°C, melt it, and then raise the water to 37°C, summing up to 390.3 kJ.
Step-by-step explanation:
The calculation involves finding the amount of energy absorbed by the climber's body when consuming snow and warming it to body temperature. To calculate the energy absorbed when a climber eats 0.75 kg of snow at -15°C and warms it to 37°C, we can use the specific heat capacity of ice, the heat of fusion for water, and the specific heat capacity of water:
The specific heat capacity of ice is approximately 2.09 kJ/kg°C.
The heat of fusion for water is 333.6 kJ/kg, which is the amount of energy required to turn ice at 0°C into water at 0°C without changing its temperature.
Water's specific heat capacity is 4.18 kJ/kg°C, which is the energy required to raise the temperature of water by 1°C.
We perform the calculation in three steps:
Raise the temperature of the snow from -15°C to 0°C:
E_1 = mass * specific heat capacity of ice * temperature change
E_1 = 0.75 kg * 2.09 kJ/kg°C * (0°C - (-15°C))
E_1 = 23.6 kJ
Melt the ice at 0°C to get water at 0°C:
E_2 = mass * heat of fusion
E_2 = 0.75 kg * 333.6 kJ/kg
E_2 = 250.2 kJ
Raise the temperature of the water from 0°C to 37°C:
E_3 = mass * specific heat capacity of water * temperature change
E_3 = 0.75 kg * 4.18 kJ/kg°C * (37°C - 0°C)
E_3 = 116.5 kJ
Summing up all the energy values:
Total energy absorbed = E_1 + E_2 + E_3
Total energy absorbed = 23.6 kJ + 250.2 kJ + 116.5 kJ
Total energy absorbed = 390.3 kJ