Ice at 0 degree celcuis is added to 200g of water initially at 70 degr...
Given data:
Initial temperature of water = 70°C
Mass of water = 200g
Mass of ice added = 50g, 80g
Final temperature of water and ice mixture after adding 50g of ice = 40°C
Final temperature of water and ice mixture after adding 80g of ice = 10°C
Calculating the heat gained by the water:
The heat gained by the water can be calculated using the equation:
Q = mcΔT
Where:
Q = heat gained by the water
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
The specific heat capacity of water is 4.18 J/g°C.
Calculation for when 50g of ice has melted:
The initial temperature of the water is 70°C and the final temperature is 40°C. Therefore, the change in temperature (ΔT) is:
ΔT = final temperature - initial temperature
ΔT = 40°C - 70°C
ΔT = -30°C
The heat gained by the water is:
Q = mcΔT
Q = (200g)(4.18 J/g°C)(-30°C)
Q = -25080 J
Calculating the heat gained by the ice:
The heat gained by the ice can be calculated using the equation:
Q = mL
Where:
Q = heat gained by the ice
m = mass of ice
L = specific latent heat of fusion of ice
Calculation for when 50g of ice has melted:
The heat gained by the ice is equal to the heat lost by the water, since the system is isolated. Therefore:
Q = -25080 J
Using the equation Q = mL, we can solve for the specific latent heat of fusion of ice (L):
-25080 J = (50g)(L)
L = -25080 J / 50g
L = -501.6 J/g
The negative sign indicates that heat is being released by the system.
Calculation for when 80g of ice has melted:
The initial temperature of the water and ice mixture is 40°C and the final temperature is 10°C. Therefore, the change in temperature (ΔT) is:
ΔT = final temperature - initial temperature
ΔT = 10°C - 40°C
ΔT = -30°C
The heat gained by the water is:
Q = mcΔT
Q = (200g)(4.18 J/g°C)(-30°C)
Q = -25080 J
The heat gained by the ice is equal to the heat lost by the water, since the system is isolated. Therefore:
Q = -25080 J
Using the equation Q = mL, we can solve for the specific latent heat of fusion of ice (L):
-25080 J = (80g)(L)
L = -25080 J / 80g
L = -313.5 J/g
Again, the negative sign indicates that heat is being released by the system.
Conclusion:
The specific latent heat of fusion of ice can be calculated to be approximately -501.6 J/g when 50g of ice is melted, and approximately -313.
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