A 3.8X10^2
B 21.3
C 17.2
D .469
E. insufficient date are given
Please walk me through it step by step.Fusion of 50 g of ice:
Moles = grams / molar mass
Moles = 50 g / 18 (g/mol) = 2.778 mol H2O
Q1 = m * L
Q1 = 2.778 mol * 6.01 kJ/mol
Q1 = 16.69 kJ
Conversion from 0ˇăC to 22ˇăC
Q2 = m * c * DT
Q2 = 2.778 mol * 75.3 J/mol * (22-0)
Q2 = 4602 J = 4.60 kJ
Total heat: Q = Q1+ Q2
Q = 16.69 + 4.60
Q = 21.29 kJ
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Solution:
B) 21.3 kJ
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Byehttp://destroyallhumans.fanfly.net/?C226...heat of fusion is the heat required to go from solid to liquid, so, you need to find the moles of water you have (use grams and molar mass for water), then multiply by 6.01 to find how many kJ it would take to chance that much ice to liquid water. using that same amount of moles, multiply 75.3 times 22 times moles to find how many JOULES it takes to raise that 0C water to 22C. Change those Joules to kJ and add it to the kJ you already found for the total.50.0 g = 2.775 mole
2.775 mole x 6.01kJ/mol = 16.68 kJ of energy to melt 50g ice
2.775 mole x 75.3 J/mol deg x 22 deg = 4597 J to heat 50g water from O to 22 deg C.
4597 j x (1 kJ /1000 J) = 4.597 kJ
16.68 kJ + 4.597 kJ = 21.3 kJ (answer B)