Question

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)

Answer 1

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...

Answer 2

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

Answer 3

That makes more sense! Give me a sec then

Answer 4

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

Answer 5

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