A sample of water has a temperature of 87.5 d/C. The volume of the sample is 430 mL. What mass of ice at 0.0 d/C will melt with this sample of water if the final temperature of water and melted ice is 0.0 d/?
Σ q = 0
(Mc∆T)water + (M∆Hfusion)ice + (Mc∆T)ice water = 0
[(430g)(1cal/g-oC)(0oC – 87.5oC)] + [(M-ice)(80cal/g)] + [(M-ice water)(1cal/goC)(0oC - 0oC)] = 0
[(430)(-87.5)] + [(M-ice)(80)] + [0] = 0
[(M-ice)(80)] = 37625 => M-ice = (37625/80)g-ice = 470g-ice
which equations did you use for this problem?
To find the mass of ice that will melt with the sample of water, we need to calculate the heat transfer between the water and the ice.
The equation to calculate the heat (Q) transferred between two substances is given by:
Q = m * c * ΔT
Where:
Q = Heat transferred (in Joules)
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance (in J/g·°C)
ΔT = Change in temperature (in °C)
In this case, the heat transferred from the water (initially at 87.5 d/C) to the ice (initially at 0.0 d/C) will result in the ice melting and reaching a final temperature of 0.0 d/C.
The specific heat capacity (c) of water is approximately 4.18 J/g·°C. Since the ice is undergoing a phase change from solid to liquid (melting), we need to account for the heat of fusion (ΔHf) of ice, which is 333.5 J/g.
Now, let's calculate the heat transferred between the water and the ice:
For the water:
Qwater = mwater * cwater * ΔTwater
For the ice:
Qice = mice * cice * ΔTice + mice * ΔHf
Since the final temperature of both the water and melted ice is 0.0 d/C, we can equate the two heat equations:
mwater * cwater * ΔTwater = mice * cice * ΔTice + mice * ΔHf
Since the water and ice are in thermal equilibrium at the end, we can assume that the heat transferred from the water is equal to the heat gained by the melted ice.
The density of water is approximately 1 g/mL. Therefore, the mass of the water is given by:
mwater = volume * density = 430 mL * 1 g/mL = 430 g
Now, rearranging the equation:
mice = (mwater * cwater * ΔTwater - mice * ΔHf) / (cice * ΔTice)
Substituting the known values:
mice = (430 g * 4.18 J/g·°C * (0.0 d/C - 87.5 d/C) - mice * 333.5 J/g) / (2.09 J/g·°C * (0.0 d/C - 0.0 d/C))
Simplifying:
mice = (430 g * 4.18 J/g·°C * (-87.5 d/C) - mice * 333.5 J/g) / 0
Since ΔTice is 0, we can ignore the term cice * ΔTice in the denominator.
Now, solving for mice:
mice = (430 g * 4.18 J/g·°C * (-87.5 d/C)) / 333.5 J/g
mice = -22.77 g
The negative sign indicates that the ice is actually gaining this mass from the water during the phase change.
Therefore, the mass of ice that will melt with this sample of water is approximately 22.77 grams.