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

Option 1 -

64.6

Option 2 -

61.7

Option 3 -

69.3

Option 4 -

63.8

0 3 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 2


    Detailed Solution:

    m(L) = m?S?(ΔT)

     ⇒ m(3.4 × 10?) = (200)(4200)(25)
    ⇒ m = 61.7

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