Heat Capacity, Change of state, Heat transfer?

Question

If you add 163.11 g if ice at 0.0 °C to 0.087 L of water at 39.9 °C and allow the water to cool to 0.0 °C, how many grams of ice will remain?

I actually have several related questions. I'm studying for an exam on Tuesday, and I'm having some problems with the following:

Q1: If you add 163.11 g if ice at 0.0 °C to 0.087 L of water at 39.9 °C and allow the water to cool to 0.0 °C, how many grams of ice will remain?

Q2:
How many kilojoules of energy would be liberated by condensing 33.22 mol of steam initially at 112.6 °C and then allowing the liquid to cool to 34.9 °C?

Q3: From the following data, calculate the total heat (in Joules) needed to convert 1.176 x 101 grams of ice at -5.859 oC to liquid water at 3.045 x 101 oC.
Melting point at 1 atm 0.0 oC
Cwater(s) = 2.09 J/g oC
Cwater(l) = 4.21 J/g oC
delta Hofus = 6.02 kJ/mol

Answer 1

First calculate the energy needed to take 0.087L (which is 87g) of water down from 39.9ºC to 0.0ºC. That is given by the specific heat of water, cp = 4.18J/gºK. The energy is

E = cp*mw*∆T

That energy comes from the melting heat of fusion cf which for water/ice is 335.5 J/g

Equate the energy values to determine how many grams of ice must melt. Subtract that from the original quantity of ice to get the remainder.

mi = E/cf = cp*mw*∆T/cf = 4.18*87*39.9/335.5 = 43.25g (of ice melts)

Remainder = 163.11g - 43.25g = 119.86g

For the second part, you need the heat of condensation for water which is 40.65 kJ/mol. You have 33.22 mole of steam, so condensing is gives up 32.22*40.65 kJ. But to condense it, you have to bring its temperature down from 112.6ºC to 100ºC, or a drop of 12.6ºC. You apply the specific heat of water vapor to cp = 37.5 J/molºK (at constant pressure, which is likely the case here). So the energy from dropping the temp of steam and condensing it is

33.22*40.65*1000 + 37.5*33.22*12.6 J

Finally, to cool the water from 100ºC to 34.9ºC, the drop in temp is 65.1ºC, and using the specific heat of water of 75.33J/molºK the total is


33.22*40.65*1000 + 37.5*33.22*12.6 + 34.9*33.22*65.1 J

Finally,

you should now know the process for solving this: look at each stage:

1) Increase the temp of the ice to the melting point (cp of ice)
2) Heat of fusion to melt the ice (∆Hf)
3) Raise the temp of the water (cp water)

Since you are given the ice in grams, but ∆Hf in moles, you need to use the fact that one mole of ice (or water) is 18g.

Answer 2

particular warmth means and warmth means are different.warmth means or thermal means is the quantity of warmth ability required to boost the temperature of the whole physique with the help of a million degree Celsius. If m is the mass of the physique and C is the particular warmth means, then:Q=mCT.As T=1degree Celsius, Q=mC×a million=mc. as a result thermal means=Mass of the physique×particular warmth. particular warmth means,in accordance to SI standards,of a substance is the quantity of warmth in joule(J) required to boost the temperature of a million kg of the substance by way of a million Kelvin.SI unit is J/kg ×ok.