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The temperature of water in equilibrium with ice can be 0 deg. C.

Molar heat capacity is specific heat capacity multiplied by molecular weight.

Specific heat capacity is the heat energy required to increase the temperature of 1 gm of water.

Suppose you have x moles of ice and y moles of water liquid in equilibrium. Then initially, the ice will fuse with all the heat available to the combination. The temperature does not rise.

Heat required for ice to become water at 0 deg C = x * latent heat of fusion

= x moles * 333.55 J /gm * 18 gm/mole = x * 6,003.9 J

Heat energy required for increase in temperature now for all x+y moles :

= (x+y)*75.327 J/degK

Then, molar heat capacity = [ 6,003.9 x + (x+y)* 75.327 J/degK * ΔT] / (x+y) ΔT

Suppose x = y , that means, ice and water are present in equal proportions, then,

molar heat capacity = heat needed to increase temperature by unit degree K

=

If ΔT = 1 deg K, then

Molar heat capacity is specific heat capacity multiplied by molecular weight.

Specific heat capacity is the heat energy required to increase the temperature of 1 gm of water.

Suppose you have x moles of ice and y moles of water liquid in equilibrium. Then initially, the ice will fuse with all the heat available to the combination. The temperature does not rise.

Heat required for ice to become water at 0 deg C = x * latent heat of fusion

= x moles * 333.55 J /gm * 18 gm/mole = x * 6,003.9 J

Heat energy required for increase in temperature now for all x+y moles :

= (x+y)*75.327 J/degK

Then, molar heat capacity = [ 6,003.9 x + (x+y)* 75.327 J/degK * ΔT] / (x+y) ΔT

Suppose x = y , that means, ice and water are present in equal proportions, then,

molar heat capacity = heat needed to increase temperature by unit degree K

=

__[ 6,003.9 + 2 * 75.327 * ΔT ] / 2 * ΔT__If ΔT = 1 deg K, then

__= 3,077.277 Joules/mol__