How many grams of ice at -14.0 degree Centigrade can be completely converted to liquid at 13.7 degree Centigrade if the available heat for this process is 5.91×10^3 kJ?
For ice, use a specific heat of 2.01J/(g.degree Centigrade}) and H2 = 6.01 kJ/mol.
Express your answer numerically in grams.
5.91 x 10^6J = [mass ice x specific heat ice x 14]+[(mass ice x heat fusion)]+[mass water x specific heat water x 13.7].
Change your 6.01 kJ/mol for heat fusion (I assume H2 stands for heat fusion) to J/g, substitute into the above and solve for grams ice. I get something like 14 kg.
To solve this problem, we need to determine the amount of heat required to change the ice from -14.0°C to 0.0°C (the melting point), and then from 0.0°C to 13.7°C.
First, let's calculate the heat required to raise the temperature of the ice from -14.0°C to 0.0°C. We can use the formula:
Q1 = (mass) x (specific heat) x (change in temperature)
Given:
Initial temperature (Ti) = -14.0°C
Final temperature (Tf) = 0.0°C
Specific heat of ice (c) = 2.01 J/(g.°C)
We want to find the heat (Q1), so rearranging the formula, we have:
Q1 = (mass) x (specific heat) x (change in temperature)
Substituting the values:
Q1 = (mass) x (2.01 J/(g.°C)) x (0.0°C - (-14.0°C))
Q1 = (mass) x (2.01 J/(g.°C)) x (14.0°C)
Next, let's calculate the heat required to melt the ice at its melting point. The heat of fusion (Hf) is defined as the amount of heat required to change one mole of a substance from a solid to a liquid phase at its melting point. For water, the heat of fusion (H2) is 6.01 kJ/mol.
Given:
Heat of fusion of ice (Hf) = 6.01 kJ/mol
To convert Hf to J/g, we need to know the molar mass of water.
The molar mass of water (H2O) = 1.007 g/mol (hydrogen) + 15.999 g/mol (oxygen) = 18.016 g/mol
Now we can calculate the heat of fusion in J/g:
Hf = (heat of fusion in kJ/mol) / (molar mass in g/mol)
Hf = (6.01 kJ/mol) / (18.016 g/mol)
Now we can calculate the heat of fusion per gram of ice:
Hf_per_gram = Hf / 1g
Finally, let's calculate the heat required to raise the temperature of the melted ice from 0.0°C to 13.7°C:
Q2 = (mass) x (specific heat of water) x (change in temperature)
Given:
Specific heat of water (c) = 4.184 J/(g.°C)
Initial temperature (Ti) = 0.0°C
Final temperature (Tf) = 13.7°C
Now we can solve for the mass of the ice (grams) using the total heat available (5.91×10^3 kJ) and the heat required to convert the ice to liquid:
Total heat available = Q1 + Q2 + heat required for fusion
Substituting the values:
5.91×10^3 kJ = Q1 + Q2 + (mass) x (Hf_per_gram)
Now, rearranging the equation to solve for the mass:
mass = (5.91×10^3 kJ - Q1 - Q2) / Hf_per_gram
After performing the above calculations, we will be able to determine the mass of ice in grams that can be completely converted to liquid at 13.7°C with the given heat available.
Please let me know if you need help with any of the specific calculations.