The change in entropy of the melting ice of 1kg at 273 kelvin is (given L= 80 cal/gm) ?

Calculate the change in entropy for the fusion of 1 mol of ice. The melting point of ice is 300K and molar enthalpy of fustion for ice = 6.0 k J mol^(-1) .

Calculate the change in entropy for the fusion of 1 mole of ice (water). The melting point of water is 273 K and molar enthalpy of funsion for water =6.0 kJ mol^(-1)

(i) Calculate the change in entropy for the fusion of 1 mole of ice. The melting point of ice is 273 K and molar enthalpy of fusion for ice = 6.0 kJ/mole. (ii) Calculate the standard of entropy change for the following reaction. {:(,H^(+)(aq), + OH(aq)to, H_(2)O(l)),(S^(@)(298 K)//JK^(-1)mol^(-1),0,-10.7,+70):}

Assertion:The change in entropy during melting of ice is negligible in comparison to change in entropy during vaporisation. Reason:The volume occupied by solid and liquids is too less in comparison to volume occupied by gas.

Calculate the entropy change in the melting of 1 kg of ice at 0^(@)C in SI units. Heat of funsion of ice = 80 cal g^(-1) .

In a insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Find the final temperature of the mixture (in kelvin.) Given, L_(fusion) = 80 cal//g = 336 J//g L_("vaporization") = 540 cal//g = 2268 J//g s_(ice) = 2100 J//kg-K = 0.5 cal//g-K and s_("water") = 4200 J//kg-K = 1cal//g-K .

A mole of steam is condensed at 100^(@) C, the water is cooled to 0^(@) C and frozen to ice . What is the difference in entropies of the steam and ice? The heat of vaporization and fusion are 540 cal" "gm^(-1) and 80 cal" "gm ^(-1) respectively . Use the average heat capacity of liquild water as 1 cal" "gm^(-1) degree^(-1) . a. Entropy change during the condensation of steam is -26.06 cal//^(@)C b. Entropy change during cooling of water from 100^(@)C "to" 0^(@)C "is" - 5.62 cal//^(@)C c. Entropy change during freezing of water at 0^(@)C "is" -5.27 cal//^(@)C d. Total entropy changwe is -36.95 cal//^(@)C

50 g of ice at 0^@C is mixed with 50 g of water at 80^@C . The final temperature of the mixture is (latent heat of fusion of ice =80 cal //g , s_(w) = 1 cal //g ^@C)

When 1 kg of ice at 0^(@)C melts to water at 0^(@)C , the resulting change in its entropy, taking latent heat of ice to be 80 cal//g is

A metal ball of specific gravity 4.5 and specific heat 0.1 cal//gm-"^(@)C is placed on a large slab of ice at 0^(@)C . Half of the ball sinks in the ice. The initial temperature of the ball is:- (Latent heat capacity of ice =80 cal//g , specific gravity of ice =0.9)