How many grams of ice at -20.1C can be completely converted to liquid at 9.9C if the available heat for this process is 4.58×103 kJ?
For ice, use a specific heat of 2.01 J/(g*C) and Heat of Fusion is = 6.01 kJ/mol.
To find the number of grams of ice that can be completely converted to liquid, we need to consider two processes: raising the temperature of the ice from -20.1°C to 0°C, and then melting the ice at 0°C.
First, we need to calculate the heat required to raise the temperature of the ice from -20.1°C to 0°C using the specific heat formula:
q1 = m * c * ΔT
where q1 is the heat absorbed, m is the mass of the ice, c is the specific heat, and ΔT is the change in temperature.
Given:
ΔT = 0°C - (-20.1°C) = 20.1°C
c = 2.01 J/(g*C)
We can convert ΔT to Kelvin by adding 273.15:
ΔT = 20.1°C + 273.15 = 293.25 K
Now we can calculate q1:
q1 = m * c * ΔT
q1 = m * 2.01 J/(g*C) * 293.25 K
Next, we need to calculate the heat required to melt the ice at 0°C using the heat of fusion formula:
q2 = m * ΔHf
where q2 is the heat absorbed, m is the mass of the ice, and ΔHf is the heat of fusion.
Given:
ΔHf = 6.01 kJ/mol
We need to convert the heat of fusion from kJ/mol to J/g:
1 kJ = 1000 J
1 mol of H2O = 18.015 g
ΔHf = (6.01 kJ/mol) * (1000 J/kJ) / (18.015 g/mol)
Now we have all the information needed to calculate the mass of ice:
q1 + q2 = 4.58×10^3 kJ
Substituting the values:
m * 2.01 J/(g*C) * 293.25 K + m * (6.01 kJ/mol) * (1000 J/kJ) / (18.015 g/mol) = 4.58×10^3 kJ
Solving this equation will give us the mass of ice that can be converted to liquid completely.
You have to heat all of the ice to 0, then melt it.
Heat=mcDeltatemp+ Hf*m
= m(2.01*(21)J/g + 6010J/mol*1mol/18g)
solve for m in grams