If steam of mass 100 g and temperature 100 °C is released on an ice s

1.	If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Answer» Data: m₁= 100 g, L₁ = 540 cal/g, T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c (water) = 1 cal/g·°C According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water: Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal Decrease in the temperature of this water to 0 °C: Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal Melting of ice: Q₃ = mL₂ = m × 80 cal/g Now, Q₁ + Q₂ = Q₃ ∴ (54000 + 10000) cal = m × 80 cal/g ∴ m = \(\frac{64000}{80}\) = 800 g 800 g of ice will melt.

If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?

Answer»

Data: m₁= 100 g, L₁ = 540 cal/g,

T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c

(water) = 1 cal/g·°C

According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:

Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal

Decrease in the temperature of this water to 0 °C:

Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal

Melting of ice: Q₃ = mL₂

= m × 80 cal/g

Now, Q₁ + Q₂ = Q₃

∴ (54000 + 10000) cal = m × 80 cal/g

∴ m = \(\frac{64000}{80}\) = 800 g

800 g of ice will melt.

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