To calculate the energy absorbed from your body, we need to account for the energy required to melt the snow and then heat the resulting water to body temperature.
First, let's calculate the energy required to melt the snow:
1. Energy to heat the snow from -10°C to 0°C:
\[ Q_1 = m \times c \times ΔT_1 \]
Where:
- \( m = 0.60 \, \text{kg} \) (mass of snow)
- \( c = 2.09 \, \text{kJ/kg}⋅^\circ \text{C} \) (specific heat capacity of ice)
- \( ΔT_1 = (0 - (-10)) \, ^\circ \text{C} = 10 \, ^\circ \text{C} \)
\[ Q_1 = 0.60 \, \text{kg} \times 2.09 \, \text{kJ/kg}⋅^\circ \text{C} \times 10 \, ^\circ \text{C} = 12.54 \, \text{kJ} \]
2. Energy to melt the snow:
\[ Q_2 = m \times L_f \]
Where:
( L_f = 334 \,{kJ/kg} \) (latent heat of fusion for ice)
[ Q_2 = 0.60 \, {kg} \times 334 \, {kJ/kg} = 200.4 \, {kJ} \]
Now, let's calculate the energy required to heat the resulting water from 0°C to 37°C:
3. Energy to heat the water from 0°C to 37°C:
\[ Q_3 = m \times c \times ΔT_2 \]
Where:
- \( ΔT_2 = (37 - 0) \, ^{C} = 37 \, ^\{C} \)
\[ Q_3 = 0.60 \, {kg} \times 4.18 \,{kJ/kg}⋅^\1 {C} \times 37 \, ^\{C} = 92.892 \, \{kJ} \]
Now, let's sum up all the energies:
\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 \]
\[ Q_{\text{total}} = 12.54 \, \text{kJ} + 200.4 \, \text{kJ} + 92.892 \, \text{kJ} \]
\[ Q_{\text{total}} = 305.832 \, \text{kJ} \]
So, the energy absorbed from your body is approximately \( 305.832 \, \text{kJ} \).
