Physics Problem about Heat Phase Change

[fireandice]

1. How many grams of ice at 0 degree Celsius must be added to 350.0 grams of water at 80.0 degree Celsius contained in an aluminum calorimeter whose mass is 150.0 grams in order to cool down the water to 30.0 degree Celsius? [Specific heat of aluminum: 910 J/kg] [Latent heat of vaporization of ice: 80 calories/gram] [Specific heat of water: 1 cal/(g x degree Celsius)]




2. a.) Q = (mass x specific heat x change in T)
b.) Q_heat_absorbed + Q_given off = 0
c.) Q = (mass x latent heat of vaporization)



3. (350 g)(1 cal/g)(30 C - 80 C) + (150 g)(910 J/kg)(1 / 4.18 J)(1 kg / 1000 g)(30 C - 80 C) + (m_ice)(80 cal/g) + (m_ice)(1 cal/g)(30 C - 0 C) = 0

The answer that I got from this equation is 173.9343193 grams. Is it what I did correct? Thanks!

 

[naresh]

Hi fireandice,

Welcome to PF. Your approach looks fine, though I haven't checked the numbers. I would not give the answer to so many significant digits, 173.9 grams should be enough -- the digits beyond that are really meaningless.

 

[thomate1]

Yes, your calculation is correct. The final answer of 173.9343193 grams is the amount of ice that needs to be added to cool down the water from 80.0 degrees Celsius to 30.0 degrees Celsius. This is because the heat absorbed by the ice (Q) is equal to the heat given off by the water and aluminum (Q_heat_absorbed + Q_given off). Using the specific heat of water and aluminum, as well as the latent heat of vaporization of ice, allows us to accurately calculate the amount of ice needed for this phase change process. Great job on your calculation!