How much it is required to convert 10 gram of ice at – 5° into stea

1.	How much it is required to convert 10 gram of ice at – 5° into steam at 100°C. Given specific heat of ice 2.1 J g-10 C-1. Latent heat of steam = 2268 J g-1 and latent hear of fusion of ice is 336 J/g. Specific heat of water = 4.2 J g-10 C-1.
Answer» Q₁ = mS_iΔT = 10 × 2.1 × 5 = 105 J Q₂ = mL_f = 10 × 336 = 3360 J Q₃ = mS_w ΔT = 10 × 4.2 × 100 = 4200 J Q₄ = mL_v =10 × 2268 = 22680 J Q = Q₁ + Q₂ + Q₃ + Q₄ = 105 + 3360 + 4200 + 22680 = 30345 J

How much it is required to convert 10 gram of ice at – 5° into steam at 100°C. Given specific heat of ice 2.1 J g-10 C-1. Latent heat of steam = 2268 J g-1 and latent hear of fusion of ice is 336 J/g. Specific heat of water = 4.2 J g-10 C-1.

Answer»

Q₁ = mS_iΔT = 10 × 2.1 × 5 = 105 J

Q₂ = mL_f = 10 × 336 = 3360 J

Q₃ = mS_w ΔT = 10 × 4.2 × 100 = 4200 J

Q₄ = mL_v =10 × 2268 = 22680 J

Q = Q₁ + Q₂ + Q₃ + Q₄

= 105 + 3360 + 4200 + 22680 = 30345 J

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How much it is required to convert 10 gram of ice at – 5° into steam at 100°C. Given specific heat of ice 2.1 J g-10 C-1. Latent heat of steam = 2268 J g-1 and latent hear of fusion of ice is 336 J/g. Specific heat of water = 4.2 J g-10 C-1.

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