Calculate the change in the temperature of the hot water. Calculate the heat los
ID: 1047880 • Letter: C
Question
Calculate the change in the temperature of the hot water. Calculate the heat lost by the hot water Heat must be expressed in kilojoules (kJ). It can be calculated by using the following equation Q = (4.18 J/g degree C) x (1 kJ/1000 J) x mass of water (g) x Delta T Calculate the volume of ice melted Calculate the mass of ice melted Remember that the density of water is 1 g/mL. Calculate the molar heat of fusion of ice, i.e.. the number of kilojoules of heat per mole ice Calculate the percent error in your determination of the value for the molar heat of fusion of Ice. The formula for Percent Error = |Experimental - Actual|/(Actual) x 100 = % Error In order to do the calculations, you assumed that all the heat lost by the hot water was absorbed by the ice, causing it to melt Was this assumption correct? Explain. Write an equation for the melting ice Include the energy term in kJ on the proper side of the equation.Explanation / Answer
1) Change in temperature of hot water = initial temperature of hot water – final temperature of water and melted ice = 55.4C – 0.5C = 54.9C (ans).
2) Volume of hot water = 30.1 mL.
Density of water is given as 1 g/mL; therefore, the mass of hot water = (30.1 mL)*(1 g/mL) = 30.1 g.
Heat lost by hot water is Q = (4.18 J/g.C)*(1 kJ/1000 J)*mass of water (g)*change in temperature (C) = (4.18 J/g.C)*(1 kJ/1000 J)*(30.1 g)*(54.9C) = 6.907 kJ 6.91 kJ (ans).
3) Volume of ice melted = (Final volume of water and melted ice) – (volume of hot water) = (48.0 mL) – (30.1 mL) = 17.9 mL (ans).
4) Mass of ice melted = (volume of ice melted)*(density of ice) = (17.9 mL)*(1 g/1 mL) = 17.9 g (ans).
5) We will assume that the heat lost by hot water was used up in melting the ice. In other words, the heat lost by hot water was absorbed by the ice and this heat caused the ice to melt.
The heat absorbed by ice = 6.91 kJ; the mass of ice melted = 17.9 g.
The molar mass of ice = 18 g/mol; therefore, the moles of ice melted = (17.9 g)*(1 mole/18 g) = 0.994 mole.
Therefore, the molar heat of fusion = heat of fusion per mole of ice = (heat absorbed by ice/moles of ice melted) = 6.91 kJ/0.994 mole = 6.9517 kJ/mol 6.95 kJ/mol (ans).
6) The accepted value of molar heat of fusion of ice = 6.02 kJ/mol; the percent error in calculation = [experimental - actual/(actual)]*100 = [(6.95 kJ/mol – 6.02 kJ/mol)/(6.02 kJ/mol)]*100 = (0.93 kJ/mol/6.02 kJ/mol)*100 = 0.15448*100 = 15.448 15.45 (ans).
7) We assumed that all the heat lost by the hot water was used up in melting the ice. Fusion or melting of ice takes place at 0C and the melted ice must be at 0C. However, note that the final temperature of the ice and water mixture is 0.5C. Therefore, a small amount of the heat lost by the hot water was used up in raising the temperature of melted ice from 0C to 0.5C. Therefore, our assumption didn’t consider this and hence the assumption was flawed.
8) The equation for melting of ice can be written as
17.9 g Ice (0C) + 6.95 kJ/mol heat -------> 17.9 g water (0C)
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