Fifty grams of ice at 0 degrees celcius is added to 200grams of water at 40 degrees celcius contained in a 100 grams copper calorimeter. If the final temperature of the mixture is 16 degrees celcius, find the heat of fusion of ice.

To find the heat of fusion of ice, we need to consider the heat gained by the water and the copper calorimeter and the heat lost by the ice.

First, let's calculate the heat gained by the water and the copper calorimeter. We will use the equation:

Q = mc∆T

Where:
Q is the heat gained or lost
m is the mass
c is the specific heat capacity
∆T is the change in temperature

For the water and the copper calorimeter, we have:
m1 = 200 g (mass of water)
c1 = 4.186 J/g°C (specific heat capacity of water)
∆T1 = (final temperature - initial temperature) = (16°C - 40°C) = -24°C

Q1 = m1 * c1 * ∆T1
= 200 g * 4.186 J/g°C * -24 °C
= -19949.76 J (Note: The negative sign indicates heat loss)

Next, let's calculate the heat lost by the ice. The heat lost by the ice equals the heat gained by the water and copper calorimeter:

Qlost = -Q1
= 19949.76 J (Note: The negative sign indicates heat loss)

Now, let's calculate the heat lost by the ice using its heat of fusion (∆Hf) and its mass:

Qlost = m2 * ∆Hf

We are given that the mass of the ice is 50 g. So, let's rearrange the equation to solve for the heat of fusion (∆Hf):

∆Hf = Qlost / m2
= 19949.76 J / 50 g
= 399.9952 J/g

So, the heat of fusion of ice is approximately 400 J/g.