Find the resulting temperature when 100 g of ice at 0oC is mixed with 1 liter of H2O at 35oC.
As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at 0°C to liquid at 0°C.
The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have
q1 + q2 = -q3
where
q1 - the heat absorbed by the solid at 0°C
q2 - the heat absorbed by the liquid at 0°C
q3 - the heat lost by the warmer water sample
The two equations that you will use are
q = m × c × ∆T, where
q - heat absorbed/lost
m - the mass of the sample
c - the specific heat of water, equal to 4.18 J/g°C
ΔT - the change in temperature, defined as final temperature minus initial temperaturea and
q = n × ∆Hfus, where
q - heat absorbed
n - the number of moles of water
ΔHfus - the molar heat of fusion of water, equal to 6.01 kJ/mol
Use water's molar mass to find how many moles of water you have in the 100.0-g sample
100.0 g × 1 mol H2O / 18.015 g = 5.551 moles H2O
So, how much heat is needed to allow the sample to go from solid at 0°C to liquid at 0°C?
q = 5.551 moles × 6.01 kJ/mole = 33.36 kJ
This means that first equation becomes
33.36 kJ + q2 = −q3
The minus sign for q3 is used because heat lost carries a negative sign.
So, if Tf is the final temperature of the water, you can say that
33.36 kJ + msample × c × ΔTsample = −mwater × c × Δ
T
water
More specifically, you have
33.36 kJ + 100.0 g × 4.18J/g°C × (Tf − 0)°C = −1000 g × 4.18 J/g°C × (Tf − 35)°C
33.36 kJ + 418 J × (Tf − 0) = −4180 J × (Tf − 35)
Convert the joules to kilojoules to get
33.36 kJ + 0.418 kJ × Tf = −4.180 kJ × (Tf − 35)
This is equivalent to
0.418 kJ × Tf + 4.180 kJ × Tf = 146.3 kJ − 33.36 kJ
Tf = 24.56°C
Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be
Tf = 25°C
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