How much boiling water at 100 degree celsius is needed to melt 2kg of ice so that the mixture which is all water is at 0 degree celsius.
Heat capacity of water-4.2 J/kg/K.
Specific latent heat of ice-336 J/g
Answers
Given : m kg of boiling water at 100°C is needed to melt 2 kg of ice so that the mixture which is all water is at 0°C.
Heat capacity of water = 4200 J/Kg/K
specific latent heat of ice = 336 J/g
To find : The amount of boiling water.
solution : heat required to melt, H₁ = 2 kg × 336 J/g
= 6.72 × 10^5 Joule
heat required to increase the temperature of molten ice, H₂ = ms∆T
= m × 4200 J/Kg/K × (100°C - 0°C)
= 4.2 m × 10^5 J
from Calorimetry,
heat lost = heat gained
⇒4.2 m × 10^5 = 6.72 × 10^5
⇒m = 6.72/4.2 = 1.6 kg
Therefore amount of boiling water is 1.6 kg needed to melt 2kg of ice so that the mixture which is all water is at 0°C.
Answer:
1.6 kg
Explanation:
The heat required to melt H₁ = 2 kg × 336 J/g
= 6.72 × 10^5 Joule
The heat required to increase the temperature of molten ice, H₂ = ms∆T
= m × 4200 J/Kg/K × (100°C - 0°C)
= 4.2 m × 10^5 J
From Calorimetry, (heat lost = heat gained )
⇒4.2 m × 10^5 = 6.72 × 10^5
⇒m = 6.72/4.2 = 1.6 kg
The answer is 1.6 kg