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प्रश्न
A vessel of negligible heat*capacity contains 40g of ice in it at 0°C, 8g of steam at 100°C is passed into the ice to melt it. Find the final temperature of the contents of the vessel.
(Specific latent heat of vaporization of steam = 2268 J/g, specific latent heat of fusion of ice = 336 J/f and specific heat capacity of water = 4.2 J/g°C)
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उत्तर
Assume, final temperature of contents = θ°C. Heat gained by ice at 0°C to water at 0°C + heat gained by water at 0°C to water at 0°C = Heat lost by steam at 100°C + Heat lost by water at 100°C to water at 0°C.
mLf + mcΔθ = mL + mcΔθ
40 × 336 + 40 × 4.2 × (θ - 0) = 8 × 2268 + 8 × 4.2 × (100 - θ)
13440 + 168 θ = 18144 + 3360 - 33.6 θ
168 θ + 33.60 = 8064 θ
201.6 θ = 8064 θ = 40°C
संबंधित प्रश्न
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