A quantity of ice at 0C was added to 33.6g of water. How much ice was added? (Heat of fusion 6.01kj/mol, specific heat 4.18J/gC)

Question

matt101 · Answer

You need more information to solve this problem. You can see that from the information given, you can add really any quantity of ice at 0C to the water...

anonymous · Answer

I left this part out ... "Added to 33.6g of water at 21C to give water at 0C."

matt101 · Answer

That makes more sense! Give me a sec then

matt101 · Answer

Consider the energy involved in the temperature change of the water. In this case, heat is being transferred from water to ice to melt the ice, so nC = -mcΔT.

n = -mcΔT/C
   = -(33.6)(4.18)(-21)/(6010)
   = 0.491 moles

0.491 * 18 = 8.84 g of ice

anonymous · Answer

Thanks. I couldn't figure out which equation to use.