0.05 kg steam at 373K and 0.45kg of ice at 253K are mixed in an insultated vessel. Find the equilibrium temperature of the mixture. Given, `L_(fusion)=80cal//g=336J//g, L_(vaporization)=540 cal//g=2268J//g,C_(ice)=2100J//KgK=0.5calg//gK and S_(water)=4200J//KgK=1cal//gK`

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 .

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 .

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 .

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)

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)

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

The amount of heat (in calories) required to convert 5g of ice at 0^(@)C to steam at 100^@C is [L_("fusion") = 80 cal g^(-1), L_("vaporization") = 540 cal g^(-1)]

The amount of heat (in calories) required to convert 5g of ice at 0^(@)C to steam at 100^@C is [L_("fusion") = 80 cal g^(-1), L_("vaporization") = 540 cal g^(-1)]