Question

Assume 12,500 J of energy is added to 2.0 moles (36 grams) of H2O as an ice sample at 0 °C. The molar heat of fusion is 6.02 kJ/mol. The specific heat of liquid water is 4.18 J/(g • °C). The molar heat of vaporization is 40.6 kJ/mol. The resulting sample contains which of the following?

A) only ice

B) ice and water

C) only water

D) water and water vapor

E) only water vapor

Answer 1

Molar Heat of fusion is the hear/energy required to melt 1 mole of solid. The molar heat of fusion for ice is 6.02 kJ/mole.

Here 12.5 kJ energy is added to 2 moles of ice (water) hence 12.04kJ energy will be utilized to melt 2 moles of ice at 0 degree celcius.......now the water is in liquid state.

Remaining energy 4600 joules.

Now specific heat is the 4.18 joules energy required to increase temperature of 1 gram of water to 1 degree celcius.

Hence, here 36 grams of water require 159.48 joules of energy to increase temperature by 1 degree.

Remaining 4600 joules will increase temperature of 36 grams of water by 30 degree celcius.

36 × 4.18 = 150.48 joules per celcius

4600 joules/150.48 joules per celcius ~ 30 degree celcius.

Hence resulting sample will be in only water.

Answer is (c) only water

Hence the