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OB: Using table B constants with energy changes in real life roblems

OB: Using table B constants with energy changes in real life roblems. (get a calculator). You have 35.0 g of ice at 273 K. If you put it into your mouth to melt it, how much energy will you impart onto that ice to convert it into liquid water without temperature change?

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OB: Using table B constants with energy changes in real life roblems

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  1. OB: Using table B constants with energy changes in real life roblems (get a calculator)

  2. You have 35.0 g of ice at 273 K. If you put it into your mouth to melt it, how much energy will you impart onto that ice to convert it into liquid water without temperature change? To do this we need a formula from table T, the heat of fusion formula. In this case we need to “un-fuse” ice into liquid water. Write the formula BIG, then under it, fill in the blanks, with units! q = mHF Whereby q is the amount of heat IN JOULES , m is the mass of H2O, and HF is the constant from table B: 334 J/g which is how much energy will it take to melt 1 g of ice with no change in temperature

  3. You have 35.0 g of ice at 273 K. How much energy to melt it without raising the temp? q = mHF

  4. q = mHF q = (35.0 g) x (334 J/g) q = 11,690 J = 11,700 Jwith 3 SF

  5. Calculate how much energy in kilojoules, it takes to freeze 1550 mL of water from liquid to ice at the freezing point. q = mHF

  6. Calculate how much energy in kilojoules, it takes to freeze 1550 mL of water from liquid to ice at the freezing point. q = mHF q = (1550 g) x (334 J/g) q = 517,700 joules q = 518,000 with 3 SF

  7. How many joules of energy will it take to vaporize 550. grams of water at the boiling point into gas at the same temperature? Hint: this time we’ll use the heat of vaporization formula with the HV constant. This is not fusing/unfusing of water at 273 K, this is the HOT phase change at 373 Kelvin

  8. How many joules of energy will it take to vaporize 550. grams of water at the boiling point into gas at the same temperature? First , write the formula, then fill it in, and cancel your units carefully. q = mHV

  9. How many joules of energy will it take to vaporize 550. grams of water at the boiling point into gas at the same temperature? q = mHV q = (550. g) x (2260 J/g) q = 1,243,000 Joules = 1,240,000 J with 3 SF

  10. When you jump into the bath water it will cool down as you absorb some energy and warm up and some energy is lost to the air. If you bath contains 625 liters of water, and it is at a comfortable 318 Kelvin, and after 15 minutes the water is cooled down to 301 Kelvin, how much energy has this water lost? Calculate joules and Kilojoules Hint: this is a change in temperature formula problem, so you cannot use the HF or HV formula, it’s time for that q = mCΔT

  11. When you jump into the bath water it will cool down as you absorb some energy and warm up and some energy is lost to the air. If you bath contains 625 liters of water, and it is at a comfortable 318 Kelvin, and after 15 minutes the water is cooled down to 301 Kelvin, how much energy has this water lost? q = mCΔT q = (625,000 g)(4.18 J/g.K)(17.0 K) q = 44,412,500 J = 44,400,000 Joules with 3 SF 44,400,000 J1 1 kJ1000J = 44,400 kJ X

  12. You know how when you go out to eat and order an appetizer and a meal and a dessert, and each part has it’s own price on the check? And then you order a soda too. You also have to pay tax on top of all this. You don’t get a little bill for each of these parts, the server adds it all up for you and you pay one price. You’re smart enough (and suave enough)to realize how simple that makes things. The same thing happens in thermochem (really).

  13. Twelve plus one equals ???

  14. What if you had say 155 g of ice at 273 K and let it sit in a room for four hours. Then the water has reached 294 K. How many joules of energy did this H2O absorb to do this? This problem is like eating out, there are smaller energy events that must be added to a whole energy total (no tax though).

  15. First, MELT the ice… q = mHF Then, warm up the water q = mCΔT Finally, add them together.

  16. What if you had say 155 g of ice at 273 K and let it sit in a room for four hours. Then the water has reached 294 K. How many joules of energy did this H2O absorb to do this? q = mHF q = (155 g)(334 J/g) = 51770 J = 51,800 J q = mCΔT q = (155 g)(4.18 J/g.K)(21.0 K) = 13605.9 J = 13,600 J ANSWER = 51,800 J + 13,600 J = 65,400 J with 3 SF Can you now convert that much energy into Calories?

  17. ANSWER = 51,800 J + 13,600 J = 65,400 J with 3 SF Convert that energy into Calories? 65,400 J1 1 cal4.18 J x = 15,645.933 cal = 15,600 cal 1 Calorie1000 cal 15,600 cal1 = 15.6 C x

  18. On a particularly bad day, 16.3 grams of steam at exactly 373 K condenses onto your finger, then cools quickly to 305 Kelvin. How much energy have you absorbed from this event? Hint: this first must condense through the hot phase change (heat of vaporization) then cool with a change in temperature formula as well. q = mHV q = mCΔT

  19. On a particularly bad day, 16.3 grams of steam at exactly 373 K condenses onto your finger, then cools quickly to 305 Kelvin. How much energy have you absorbed from this event? q = mHV q = (16.3 g)(2260 J/g) = 36,838 J = 36,800 J with 3 SF q = mCΔT q = (16.3 g)(4.18 J/g.K)(68.0 K) = 4633.112 J = 4630 J ANSWER: 36,800 J + 4630 J = 41430 J = 41,400 J with 3 SF

  20. Do Thermochem HW #1(handout) Read the diary, do not wait until you’re too busy. Read it tonight!

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