Heat Problems

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Heat Basics HEAT (cal) = SPECIFIC HEAT (cal/g-°C) x MASS (g) x CHANGE IN TEMPERATURE (°C)

  1. How many calories are released when 80.0 g of water cools from 37.5 °C to 22.5 °C?
  2. What is the final temperature of the water when 0.500 kcal are added to 135.8 g of water at 14.8 °C?
  3. A bar of lead (specific heat 0.030 cal/g-°C) is cooled from 148.0 to 20.4 °C. If this releases 324.8 calories of heat, what is the bar's mass?
  4. A 15.0 kg ball of lead (specific heat 0.030 cal/g-°C) is cooled to 21.5 °C with a loss of 0.382 kcal of heat energy. What was the original temperature of the lead?How many calories are needed to melt 150.0 g of ice at 0.0°C? The heat of fusion of ice is 1.44 kcal/mole.
  5. How many grams of water can be condensed at 100.0 °C by the removal of 500.0 cal of heat? the heat of vaporization of water is 9.7 kcal/mole.
  6. How many calories of heat are absorbed when 15.0 g of lead at its melting point is converted completely to a liquid at the same temperature? The heat of fusion of lead is 5.9 cal/g.
  7. Calculate the heat of fusion of copper if it requires 1230 cal of heat to melt 38.4 g of copper at its melting point.

1. -1200 cal

5. 12.0 kcal

2. 18.5 °C

6. 0.928 g

3. 85 g

7. 89 cal

4. 22.3 °C

8. 2.03 kcal/mole
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Changes in State

  1. How much energy is needed to change a 25.0 g ice cube at -10.0 °C to steam at 110.0°C? The specific heat of steam is 0.50 cal/g-°C.
  2. An 850.0 mg sample of lead is kept in a basement at 10.0 °C . If 25.0 cal of heat energy are supplied to it from a nearby furnace, will the sample melt completely? Lead melts at 327°C.
  3. 0.500 kcal of heat energy are supplied to a 50.00 g block of lead at room temperature (20.0°C). What temperature and state will the lead reach? Lead melts at 327°C and boils at 1750°C.
  4. A 2.50 kg sample of molten silver at its melting point of 961°C is placed in a container of liquid nitrogen (-195°C). How much energy must be removed to bring the solid silver down to the temperature of the N2 (l)?
  5. Suppose there is a beaker containing 25.0 mL of methanol in front of you in the lab, which is at about 20°C. If you add 500 cal of heat energy, will it boil? The bp of methanol is 65°C. Assume 60% of the energy is "wasted" in heating the beaker, and only 40% goes into heating the methanol. The density of methanol is 0.80 g/mL, and its formula is CH3OH.

ANSWERS:

(1) warm ice = (0.50 cal/g-°C)(25.0 g)(10°C) = 125 cal = 0.125 kcal

melt ice = (25.0 g)(1 mol/18.0 g)(1.44 kcal/1 mol) = 2.00 kcal

heat water = (1.00 cal/g-°C)(25.0 g)(100°C) = 2500 cal = 2.50 kcal

boil water = (25.0 g)(1 mol/18.0 g)(9.7 kcal/1 mol) = 13.5 kcal

heat steam = (0.50 cal/g-°C)(25.0 g)(10°C) = 125 cal = 0.125 kcal

TOTAL = 18.3 kcal

(2) to heat solid Pb = (0.031 cal/g-°C)(0.850 g)(327 - 10°C) = 8.35 cal neded

to melt Pb = (0.850 g)(1 mol/207.2 g)(1.14 kcal/1 mol) = 0.00468 kcal = 4.68 cal

total needed to melt Pb = 8.35 cal + 4.68 cal = 13.0 cal < 25.0 cal available

YES, the Pb will melt completely.

(3) heat solid Pb = (0.031 cal/g-°C)(50.00 g)(327 - 20°C) = 476 cal = 0.476 kcal

With 0.500 kcal available, the lead will at least start to melt.

to melt Pb = (50.0 g)(1 mol/207.2 g)(1.14 kcal/1 mol) = 0.2755 kcal

This is much greater than the 0.500 kcal available, so … The lead will be partially molten at 327°C.

A better answer is to determine just how much of the lead has melted. There is 0.500 - 0.476 = 0.024 kcal available for melting the Pb. So …

(0.024 kcal)(1 mol/1,14 kcal)(207.2 g)/1 mol) = 4.4 g Pb There will be a mix of 4.4 g molten lead and 45.6 g solid lead at 327°C.

(4) freeze silver = (2500 g)(1 mol/107.9 g)(2.84 kcal/1 mol) = 65.8 kcal

cool solid = (0.61 cal/g-°C)(2500 g)[961-(-195°C)] = 1,763,000 cal = 1763 cal

TOTAL = 1830 kcal

(5) 25.0 ml x (0.80 g/1 mL) = 20.0 g methanol

40% of 500 cal = 200 cal available to boil methanol

heat methanol = (0.057 cal/g-°C)(20.0 g)(65-20°C) = 51.3 cal

boil methanol = (20.0 g)(1 mol/32.0 g)(8.01 kcal/1 mol) = 5.05 kcal = 5050 cal

There is enough energy to start to boil the methanol, but not enough to boil all of it,because 51.3 cal < 200 < 5050 cal.

Again, you can improve this answer by specifying how much of the methanol will boil off. There is 200 - 51.3 = 149. calories available for boiling. So … (0.149 kcal)(1 mol/8.01 kcal)(32.0 g/1 mol) = 0.595 g. CH3OH boils off

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Heat Equilibria

  1. A 25.0 g cube of aluminum at 150.0 °C is placed in 100.0 g of water at 25.0 °C. What's the final temperature of the mixture?
  2. A 200.0 g bar of silver that has been heated to 125°C is placed in 150.0 g of water. The water temperature rises from 20.0 °C to 27.3 °C. What is the specific heat of silver?
  3. 250 g. of water warms from 28.4 to 33.9 °C when solid lead at 200.0 °C is added to it. What is the mass of the lead?
  4. A 45.0 g cube of iron heated to 142.5 °C is placed in a beaker of water, which then warms from 21.5 to 24.3 °C. What is the mass and volume of the water?
  5. When 200.0 g of iron at 162.5 °C was added to 300.0 mL of water, the iron cooled to 36.5 °C. What was the original temperature of the water?
  6. In an experiment to confirm the specific heat of water, a student adds 85.0 g of lead shot at 100 °C to a calorimeter containing 250.0 g of water at 18.6 °C. The equilibrium temperature is 19.5 °C. What is the student's value for the specific heat of water? How far off was s/he?
Answers:

  1. 1. (0.22 cal/g-C)(25.0 g)(Tf-150.0 °C) = (-1)(1 cal/g-°C)(100.0 g)(Tf - 25.0 °C)

    Tfinal = 32 °C

  2. (sp. heat)(200.0 g)(27.3-125 °C) = (-1)(1 cal/g-°C)(150.0 g)(27.3-20.0 °C)

    specific heat = 0.056 cal/g-°C

  3. (0.031 cal/g-°C)(m)(33.9-200.0 °C) = (-1)(1 cal/g-°C)(250.0 g)(33.9- 28.4 °C)

    mass = 270 g

  4. (0.031 cal/g-°C)(85.0 g)(19.5-100.0 °C) = (-1)(sp. heat)(250.0 g)(19.5-18.6°C)

    specific heat = 0.9 cal/g-°C; off by 0.1 cal/g-°C or 10%

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