The specific heat of water =4200Jkg-1K-1 and the latent heat of ice =3.4×105Jkg-1 . 100 grams of ice at 0C is placed in 200g of water at 25C . The amount of ice that will melt as the temperature of water reaches 0C is close to (in grams):

Option 1 -

 64.6

Option 2 -

61.7      

Option 3 -

69.3

Option 4 -

63.8

0 5 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    m (L)=m1S1 (ΔT)

    m3.4×105= (200) (4200) (25)

    m=61.7

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