How much ice (at 0oC) must be added to 1.0 kg of water at 100oC to end up with all the liquid at 20oC?

Answer 1

The ice must melt and heat up to 20 C.
The water must cool from 100 to 20 C.
Both heating values must be equal in magnitude.

Q = m*L + m*Cp*20
Q = 1*Cp*(100-20)

So m = 80Cp / (L + 20Cp)
Just find the values and plug them in.

If you wish to be more accurate, you can account for changes in Cp by considering enthalpy instead.
m = (h100 - h20) / (L0 + h20 - h0)

Answer 2

You need this data:
Heat of melting of ice
Specific heat of water [you will probably get away with assuming it is constant over 0 - 100C]

Then calculate how much heat a kg of ice absorbs going from ice at 0C to water at 20C, call it Ha

Calculate how much heat 1 kg water gives up going from 100 C to 20 C call it Hb

Mass of ice required = Hb/Ha

Answer 3

1.0 kg x (100 oC - 20 oC) = 80 kcal, this must be equal to the number of calories it needs to turn ice from 0 degrees Celsius to 20 degrees Celsius.Thus,

80 kcal = (kg of ice) x (20 oC - 0 oC)
Kg of ice = 4 kg

Answer 4

let ci = specific heat of ice, mi = mass of ice, cw = specific heat of water, mw = heat of water and H be heat of fusion for water.

Heat of fusion is energy needed to melt a given amount of ice into water at 0 deg C. Now the ice melting draws heat out of the water by an amount.

mi*H = mw*cw*DT' where DT' is the temperature change from 100 C to whatever the temperature ends up after the ices melts, T1. So

DT' = mi*H/(mw*cw) = (100 -T1)

Now we have a mass of mi water at 0 C and a mass mw of water at T1. The mass mw will give up heat to raise the temperature of mass mi from 0 C to 20 C. So

mi*cw*20 = mw*cw*(T1-20) = mw*cw*(100 -mi*H/(mw*cw) -20)

mi*cw*20 = mw*cw*(80 -mi*H/(mw*cw))

mi*(20 +H/(mw*cw)) = 80*mw

mi = 80*mw/(20+H/(mw*cw))

I don't have teh values for H handy - you can get them from your text or the internet.