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

Question

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

anonymous · Answer

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.

anonymous · Answer

thank you! that really helped

anonymous · Answer

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