When steam condenses to liquid water. 2.26 kJ of heat is released per gram. The
ID: 996421 • Letter: W
Question
When steam condenses to liquid water. 2.26 kJ of heat is released per gram. The heat from 168 g of steam is used to heat a room containing 6.44 times 10^4 g of air (20 ft times 12 ft times 8 ft). The specific heat of air at normal pressure is 1.015 J/(g middot degree C). What is the change in air temperature, assuming the heat from the steam is all absorbed by air? When ice at 0degree C melts to liquid water at 0 degree C. it absorbs 0.334 kJ of heat per gram. Suppose the heat needed to melt 31.5 g of ice is absorbed from the water contained in a glass. If this water has a mass of 0.210 kg and a temperature of 21.0 degree C. what is the final temperature of the water? (Note that you will also have 31.5 g of water at 0 degree C from the ice.) When 15.3 g of sodium nitrate. NaNO_3. was dissolved in water in a calorimeter, the temperature fell from 25.00 degree C to 21.56 degree C. If the heat capacity of the solution and the calorimeter is 1071 J/degree C. what is the enthalpy change when 1 mol of sodium nitrate dissolves in water? The solution process is NaNO_3(s) rightarrow Na^+(aq) + NO_3^- (aq); delta H = ?Explanation / Answer
6.65 )
Heat of condensation of steam = 2.26 KJ / gram
Heat used = 2.26 X 168 = 379.68 KJ
amount of air = 6.44 X 10^4 grams
Specific heat of air = 1.015
so heat absrobed by air = Mass of air X specific heat X change in temperature
379.68 X 10^3 Joules = 6.44 X 10^4 grams X 1.015 X change in temperature
Change in temeprature = 379.68 X 10^3 / 6.44 X 10^4 X 1.015 = 5.81 0C
66) Ice absrobs heat = 0.334 KJ /gram
Heat needed to melt 31.5 g of ice = 0.334 X 31.5KJ = 10.512 KJ
Specific heat of water = 4.184 J / g C
Mass of water = 31.5 g + 210 g
Heat Loss by water = Mass of water X specific heat of water X change in tempeature
10.512 X 10^3 Joules = 241.5 X 4.184 X change in temperature
Change in temperature = 10.403 0C
Initial temperature = 21 0C
Final temperature = 21 -10.4 = 10.60C
67) The change in temperature = 25-21.56 = 3.44 0C
Mass of sodium nitrate = 15.3 grams
Heat capacity of calorimeter = 1071 J / 0C
Heat loss by calorimeter = Heat capacity X change in temperature = 1071 X 3.44 Joules = 3684.24 Joules
This heat is associated with 15.3 grams of NaNO3
Molecular weight of NaNO3 = 23 + 14 + 48 = 85
so heat associated with 1 gram = 3684.24 / 15.3
So heat associated with 85 grams = 3684.24 X 85 / 15.3 = 20468 Joules
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