Physics Problem about Heat Phase Change

  • Thread starter fireandice
  • Start date
  • #1
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!
 

Answers and Replies

  • #2
166
0
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.
 

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