How much steam at 100° C will just melt 3200 g of ice at -10° C ? (Specific heat of ice =0.5 cal / g ° C, specific heat of water =1 cal / g ° C, latent heat of fusion of ice =80 cal / g, and latent heat of vaporisation of water =540 cal / g )

Question

How much steam at 100° C will just melt 3200 g of ice at -10° C ? (Specific heat of ice =0.5 cal / g ° C, specific heat of water =1 cal / g ° C, latent heat of fusion of ice =80 cal / g, and latent heat of vaporisation of water =540 cal / g ) (A) 425 g (B) 525 g (C) 625 g (D) 725 g

Tardigrade Admin · Accepted Answer

Let m grams be the mass of the steam. Heat lost by the steam =m × L + m × 1 ×(100-0) =m × 540+100 m=640 m Heat gained by ice =mi × s × Δ T+mi L =3200 × 0.5 ×[0-(-10)]+3200 × 80 =272000 cal According to the principle of calorimetry 640 m=272000 or m=425 g

Q. How much steam at 100∘C will just melt 3200g of ice at −10∘C? (Specific heat of ice =0.5cal/g∘C, specific heat of water =1cal/g∘C, latent heat of fusion of ice =80cal/g, and latent heat of vaporisation of water =540cal/g )

425 g

525 g

625 g

725 g

Let m grams be the mass of the steam. Heat lost by the steam =m×L+m×1×(100−0) =m×540+100m=640m Heat gained by ice =mi​×s×ΔT+mi​L =3200×0.5×[0−(−10)]+3200×80 =272000cal According to the principle of calorimetry 640m=272000 or m=425g