How many grams of ice at -14.3∘C can be completely converted to liquid at 24.8∘C if the available heat for this process is 4.75×103kJ ?
q1 = heat to raise ice from -14.3 to zero.
q1 = mass ice x specific heat ice x (Tfinal-T(initial)
q2 = heat to melt ice
q2 = mass ice x heat fusion
q3 = heat to raise T liquid H2O from zero to 24.8
q3 = mass water x specific heat H2o x (Tfinal-Tinitial)
Then q1 + q2 + q3 = 4.75E3 kJ.
I would convert kJ to J, then sum q1 + q2 + q3 and solve for mass ice/water. I did a quick estimate and it's approx 1E4 g ice if I punched in all of the right numbers.
You have to have the following values. These can be found in google or in textbooks:
*specific heat capacity of ice (c,ice) in J/g-K
*latent heat of fusion for water (Hf) in J/g
*specific heat capacity of water (c,water) in J/g-K
Then substitute them in the summation of all energies involved:
(Q,ice) + (Q,phase change) + (Q,water) = 4.75 x 10^3 kJ (1000 J /kJ)
Note that the freezing point of water is at 0 deg Celsius.
We're solving for m:
m(c,ice)(0 - T1) + m(Hf) + m(c,water)(T2 - 0) = 4.75 x 10^6
m(c,ice)(0 - (-14.3)) + m(Hf) + m(c,water)(24.8 - 0) = 4.75 x 10^6
You have values for c,ice, Hf and c,water so you can calculate for m.
Hope this helps~ `u`
To find out how many grams of ice can be completely converted to liquid, we first need to calculate the heat required for the phase change from solid (ice) to liquid (water).
The heat required (q) for a phase change can be calculated using the formula:
q = m * ΔH
Where:
q = heat required (in Joules)
m = mass (in grams)
ΔH = heat of fusion (or heat of melting) for the substance (in J/g)
The heat of fusion for ice is 334 J/g.
Now, let's calculate the heat required to convert the ice to water:
q = m * ΔH
4750 J = m * 334 J/g
To eliminate the J units, we divide both sides of the equation by 334 J/g:
4750 J / 334 J/g = m
m ≈ 14.22 g
Therefore, approximately 14.22 grams of ice can be completely converted to liquid at 24.8∘C with the available heat of 4.75×103 kJ.