Best Answer:
Specific heat capacity, ice: 2108 J/kg-C`

Specific heat capacity, water: 4187 J/kg-C`

latent heat of fusion for ice is 3.35*10^5 J/kg

Energy required for the temperature rise form -8 degree celsius to 0 degree celsius is;

E=mc(<>T)

=0.80 x 2108 x 8

=13491.20J

E=mc(<>T) where E=energy, m=mass, c=Specific heat capacity of ice, <>T= change in temperature

--------------------------------------...

Energy required for the ice-to-water conversion is;

E=mL

=0.80 x 3.35*10^5

=268000J

E=mL where E=energy, m=mass, L= latent heat of fusion of ice

--------------------------------------...

Energy required for the temperature rise form 0 degree celsius to 37 degree celsius is;

E=mc(<>T)

=0.80 x 4187 x 37

=123935.20J

E=mc(<>T) where E=energy, m=mass, c=Specific heat capacity of water, <>T= change in temperature

--------------------------------------...

Therefore, now add up all the energy(s) to calculate the total energy required for the whole process;

Total energy = 13491.20 + 268000 + 123935.20

= 405426.40J

= 405.43 KJ

Therefor the energy absorbed from your body if you eat 0.80kg of -8 degree celsius snow which your body warms to body temperature of 37 degree celsius is 405.43 KJ

N.B Use the values of Specific heat capacity of water, ice and specific latent heat of fusion of ice give in your book/question.

Source(s):