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Chemistry 8 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

thank you! that really helped

OpenStudy (anonymous):

No problem just remember to use the specific heat for ice and then for the water because they are different.

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