Thermochemistry help?

Question

How much ice must be added to 540g of water at 25C to cool the water to 0 C ?

Answer 1

The total heat energy Q = m * Cp * ΔT. Now you have to apply this on both the water streams

m1 * Cp1 * ΔT1 = m2 * Cp2 * ΔT2

m1 = 540/1000 = 0.54 kg

Cp1 = 4.184 KJ/ kg C (Specific heat for water)

ΔT1 = Temperature difference (25 - 0 C) = 25 C

Now since in the other stream you have used ice which will give away its latent heat and not specific heat, you need to use its latent heat (lamda)

m2 = mass of ice = ?

ΔT2 = temperature difference (here zero because only latent heat will be used, this is assuming the ice is at 0 degree C)

Lamda = latent heat of fusion of water = 334 KJ/kg

Now the equation is

m1 * Cp1 * ΔT1 = m2 * lamda

Substituting

0.54 * 4.184 * 25 = m2 * 334

Solving for m2

m2 = (0.54 * 4.184 * 25) / 334 = 0.169 kg

The amount of ice required is 0.169 kg (169 gram)

Answer 2

Do the heat balance...  what is the starting temp of the ice, startng temp of liqid water, Cp for ice, latent heat of melting, and Cp of liquid water.  If it does not define what the starting temperatur of the ice is, just for novelty, assume the solid ice is at absolute zero and assume the specific heat of ice (Cp) is constant from absolute zero to melting point (which is not true but is a good simplifying assumption).  good luck