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From the data below, calculate the total heat (in J) needed to convert 12.0 g of ice at -5.0 ºC to liquid water at 0.5 ºC: ΔHfus= 6.02 kJ/mol, cliquid = 4.18 J/g ºC, csolid = 2.09 J/g ºC
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use this for ice from -5 degree to 0 degree \[q=C_s*m*\Delta T\] then use \[12.0g H_2O (\frac{1 mole H_2O}{18.02 g H_2O})(\frac{6.02 KJ}{1 mole H_2O})= 4.01 KJ\] This is for the phase change the use the top equation for 0 degree to 5 degree, and add up all the different parts and you will have your total heat.
thank you! that really helped
No problem just remember to use the specific heat for ice and then for the water because they are different.
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