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.
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 ?
2 answers
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`
*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`
1 answer