Question

In an experiment to determine the enthalpy of fusion of ice, the following data was collected:

Initial mass of water in the calorimeter = 70.89g

Initial temperature of the water in the calorimeter = 17.6 degrees C

Final mass of temperature in the calorimeter = 0.0 degrees C

Please calculate:

a) the mass of the ice that melted

b) the number of mol of ice that melted

c) the change in temperature of the initial mass of water in the calorimeter

d) the energy released by the initial mass of water (the specific heat capacity of water is 4.184 J.g/c)

e) the energy that was absorbed by the ice that melted

f) the energy required to melt 1 mol of ice

Answer #1

You have not provided complete data.

Final mass of water is not given.

Assume that initial mass of water is m_{1} and final
mass of water is m_{2}.

Initial temp of water is T_{1} and final is
T_{2}.

So,

Mass of ice that melted = m' = m_{2}-m_{1}
grams

Change in temperature, dT = T_{1}-T_{2}.

Heat absorbed by melting of ice = m'*L

Heat released by water = m_{1}*C*dT

Here, C = heat capacity of water = 4.18 J/(g.^{0}C)

Balancing the heat released by the heat absorbed, we get:

m_{1}*C*dT = m'*L

Putting values:

m_{1}*4.18*(T_{1}-T_{2}) =
(m_{2}-m_{1})*L

Solve for L from the above equation. This is the enthalpy of fusion of ice.

Hope this helps !

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