Latent heat of fusion of ice =80calg−1 =80×4.2J×103kg−1 =336×103Jkg−1

Was this answer helpful?

4

Similar Questions

Q1

Assertion :Latent heat of fusion of ice is 336000Jkg−1. Reason: Latent heat refers to change of state without any change in temperature.

View Solution

Q2

400 g of ice at 253 K is mixed with 0.05 kg of steam at 100oC. Latent heat of vaporisation of steam =540calg−1. Latent heat of fusion of ice =80calg−1. Specific heat of ice =0.5calg−1oC−1. Find the resultant temperature of the mixture.

View Solution

Q3

Work done in converting one gram of ice at −10∘C into steam at 100∘C is -
[Given, specific heat of ice ci=0.5cal g−1∘C−1, specific heat of water cw=1cal g−1∘C−1, latent heat of fusion of ice Lf=80cal/g, & latent heat of vaporization of steam Lv=540cal/g]

View Solution

Q4

Calculate the amount of heat reqiured to convert 1kg of ice at −10∘C into steam at 100∘C. Specific heat capacity of ice is=2100Jkg−1K−1, latent heat of fusion of ice =3.36×105Jkg−1, specific heat capacity of water =4200Jkg−1K−1 and latent heat of vapourization of water=2.25×106Jkg−1.

View Solution

Q5

What is the entropy change in J.K−1 during the melting of 27.3 grams of ice at 0∘C? (Latent heat of fusion of ice =330J.g−1)