Ice at 0°C is added to 200gms of water initially at 70°C in a vacuum flask. When 50gms of ice has been added and has all melted the temperature of the flask and the contents become 40°C. When 80gms of ice has been added and has all melted the temperature becomes 10°C. Neglecting heat loss to the surroundings calculate latent heat of fusion of ice
To calculate the latent heat of fusion of ice, we need to use the principle of conservation of energy. We know that the heat gained by melting the ice is equal to the heat lost by cooling the water.
Let's break down the problem step by step:
1. Find the heat lost by cooling the water from 70°C to 40°C:
- Mass of water (m₁) = 200 g
- Initial temperature (T₁) = 70°C
- Final temperature (T₂) = 40°C
- Specific heat capacity of water (c) = 4.184 J/g°C (approx.)
The heat lost by cooling the water can be calculated using the formula:
Q₁ = m₁ * c * (T₁ - T₂)
2. Find the heat gained by melting 50 g of ice:
- Mass of ice (m₂) = 50 g
- Latent heat of fusion of ice (L) = ?
The heat gained by melting the ice can be calculated using the formula:
Q₂ = m₂ * L
3. Set Q₁ equal to Q₂ and solve for L:
m₁ * c * (T₁ - T₂) = m₂ * L
4. Repeat steps 1-3 with different values:
- Find the heat lost by cooling the water from 40°C to 10°C.
- Find the heat gained by melting 80 g of ice.
- Set the two equal and solve for L.
By following these steps, you can calculate the latent heat of fusion of ice.
To calculate the latent heat of fusion of ice (L), we need to use the principle of conservation of energy.
Assuming no heat is lost to the surroundings and no phase changes occur other than melting of the ice, we can break down the problem into two steps:
Step 1: Heat gained by the ice to reach 0°C.
Step 2: Heat gained by the ice at 0°C to melt completely.
Let's calculate the heat gained by the ice in Step 1:
Heat gained by the ice in Step 1 = mass of ice * specific heat capacity of ice * change in temperature
Given:
Mass of ice in Step 1 = 50 g
Specific heat capacity of ice = 2.09 J/g°C
Change in temperature = 0°C - (-10°C) = 10°C
Heat gained by the ice in Step 1 = 50 g * 2.09 J/g°C * 10°C
= 1045 J
Now, let's calculate the heat gained by the ice in Step 2:
Heat gained by the ice in Step 2 = mass of ice * latent heat of fusion of ice
Given:
Mass of ice in Step 2 = 80 g
Heat gained by the ice in Step 2 = 80 g * L
The total heat gained by the ice is equal to the heat lost by the water:
Heat lost by the water = mass of water * specific heat capacity of water * change in temperature
Given:
Mass of water = 200 g
Specific heat capacity of water = 4.18 J/g°C
Change in temperature = 70°C - 40°C = 30°C
Heat lost by the water = 200 g * 4.18 J/g°C * 30°C
= 25080 J
Since there is no heat loss to the surroundings and the total heat gained by the ice is equal to the heat lost by the water, we can equate the two:
1045 J + 80 g * L = 25080 J
Rearranging the equation:
80 g * L = 25080 J - 1045 J
L = (25080 J - 1045 J) / 80 g
L = 30035 J / 80 g
Calculating the value of L:
L = 375.4375 J/g
Therefore, the latent heat of fusion of ice is approximately 375.44 J/g.
the water in the flask loses heat to the ice
... (mass of water) * (specific heat of water) * (temperature change)
the ice melts , and the resulting water is heated to the final flask temperature
... (mass of ice) * (latent heat of fusion of ice)
... plus ... (mass of water) * (specific heat of water) * (temperature change)