The latent heat of fusion of water at 0 °C is 6.025 kJ mol'' and the molar heat capacities (C ) of water and ice are 75.3 and 37.7JK'mol', respectively. Calculate
AH for the freezing of 1 mol of supercooled water at -10.0°C.

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of . Then:

Finally, the amount of heat released from water is now computed by replacing variables:

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j