Ice at 0 degree celcuis is added to 200g of water initially at 70 degree celcuis in a vacuum flask.When 50g of ice has been added and has all melted the temperature of the flask and contents is 40 degree celcuis.When a further 80g of ice has been added and has all melted the temperature of the whole becomes 10 degree celcuis.Calculate the specific latent heat of fusion of ice, neglecting any heat lost to the surroundings.

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To calculate the specific latent heat of fusion of ice, we need to understand the concept of heat transfer in this scenario.

Initially, we have 200g of water at 70 degrees Celsius and some ice at 0 degrees Celsius in a vacuum flask. As the ice melts, it absorbs heat from the water, leading to a decrease in the water's temperature. When all the ice has melted, the temperature of the flask and its contents reaches 40 degrees Celsius.

Let's determine the amount of heat transferred when the first 50g of ice melts:

The heat gained by the ice can be calculated using the equation:

Q1 = ml1,

where Q1 is the heat gained by the ice, m is the mass of the ice, and l1 is the specific latent heat of fusion of ice.

Assuming the initial temperature of the water is constant at 70 degrees Celsius:

Q1 = ml1 = 50g x l1

The heat lost by the water can be calculated using the equation:

Q2 = mcΔT,

where Q2 is the heat lost by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Q2 = mcΔT = 200g x 4.18 J/g°C x (70°C - 0°C)

Since the heat gained by the ice Q1 is equal to the heat lost by the water Q2, we can equate the equations:

Q1 = Q2
50g x l1 = 200g x 4.18 J/g°C x (70°C - 0°C)

Now, let's determine the amount of heat transferred when the next 80g of ice melts:

This time, the final temperature is 10 degrees Celsius. Similar to before, we can set up an equation:

Q3 = ml2,

where Q3 is the heat gained by the ice, m is the mass of the ice, and l2 is the specific latent heat of fusion of ice.

Assuming the final temperature of the water is constant at 40 degrees Celsius:

Q3 = ml2 = 80g x l2

The heat lost by the water can be calculated using the equation:

Q4 = mcΔT = 200g x 4.18 J/g°C x (40°C - 10°C)

Since Q3 is equal to Q4, we can equate the equations:

Q3 = Q4
80g x l2 = 200g x 4.18 J/g°C x (40°C - 10°C)

Now, let's solve these two equations to find the specific latent heat of fusion of ice (l1 and l2):

50 x l1 = 200 x 4.18 x 70
80 x l2 = 200 x 4.18 x 30

Calculate the values on both sides of each equation to solve for l1 and l2:

50 x l1 = 58040
80 x l2 = 50160

Simplifying:

l1 = 58040 / 50
l2 = 50160 / 80

Performing the calculations:

l1 = 1160.8 J/g
l2 = 627.0 J/g

Now, to find the average specific latent heat of fusion of ice:

lave = (l1 + l2) / 2

lave = (1160.8 + 627.0) / 2

lave ≈ 893.9 J/g

Therefore, the specific latent heat of fusion of ice, neglecting any heat lost to the surroundings, is approximately 893.9 J/g.

To calculate the specific latent heat of fusion of ice, we need to use the principle of energy conservation. Let's consider the energy transfers in each step:

Step 1: Ice at 0°C is added to the water at 70°C until all 50g of ice melt.
In this step, the heat lost by the water is equal to the heat gained by the ice.

Q1 = m1 * c * (T2 - T1)

Where:
Q1 = heat lost by hot water, which is equal to the heat gained by the ice
m1 = mass of hot water = 200g
c = specific heat capacity of water = 4.18 J/g°C
T1 = initial temperature of hot water = 70°C
T2 = final temperature after ice melts = 0°C

Q1 = 200g * 4.18 J/g°C * (0 - 70)°C

Step 2: The resulting water at 0°C is further cooled down to 40°C after adding 80g of ice.
In this step, the heat lost by the water is equal to the heat gained by the ice.

Q2 = m2 * c * (T3 - T2)

Where:
Q2 = heat lost by the water, which is equal to the heat gained by the ice
m2 = mass of water + melted ice from step 1 = (200g + 50g = 250g)
T3 = final temperature after ice melts = 40°C
T2 = initial temperature of the water after the ice from step 1 melts = 0°C

Q2 = 250g * 4.18 J/g°C * (40 - 0)°C

Step 3: The resulting water at 40°C is further cooled down to 10°C after adding 80g of ice.
Again, the heat lost by the water is equal to the heat gained by the ice.

Q3 = m3 * c * (T4 - T3)

Where:
Q3 = heat lost by the water, which is equal to the heat gained by the ice
m3 = mass of water + melted ice from step 2 = (250g + 80g = 330g)
T4 = final temperature after ice melts = 10°C
T3 = initial temperature of the water after the ice from step 2 melts = 40°C

Q3 = 330g * 4.18 J/g°C * (10 - 40)°C

Now, the total heat gained by the ice can be calculated by adding up the heat transferred in each step:

Q_total = Q1 + Q2 + Q3

Finally, we can calculate the specific latent heat of fusion of ice (L):

L = Q_total / m_ice

Where:
m_ice = mass of ice = (50g + 80g = 130g)

Substituting the values and calculating will give you the specific latent heat of fusion of ice.