Q = mc T f T i ) Q = mc T)


 Eugene Harmon
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1 Problem Solving with Heat Heat is quite a complex concept. Heat can be effected by how much of the substance there is, what temperature the substance is at, and what the substance is. We need a unit define and quantify heat. It is a unit that you are very familiar with already; the calorie. Calories A calorie is a unit of energy, of heat. With every substance having a different specific heat capacity, it can be difficult to come up with a unit that describes all energy. It was decided to pick a common substance to act as a standard against which all other substances would be measured. The common substance picked was water. We know that mass, specific heat, and temperature all effect the energy of a substance so to keep the numbers very simple: a calorie was defined as the amount of energy needed to raise 1 grams of water by 1 o C Do you notice that all three aspects of heat are included; mass, temperature, and identity? Thus, the specific heat capacity of water (as indicated in the previous section) is defined as 1 cal/g o C. So if 1 calorie of heat is applied to 1 gram of water at 20 o C, the water increases to 21 o C. If 10 calories of heat are applied to 1 gram of water at 20 o C, the water increases to 30 o C. If 10 calories of heat are applied to 10 grams of water at 20 o C, the water increase to 21 o C! Let s put all this into a simple equation which includes all of our variables: Heat Mass Specific heat capacity Q = mc T f T i ) Q = mc T) Change in temperature The heat quantity is Q, the mass is m, the specific heat capacity is c, and the final temperature is T f while the initial temperature is T i.. Another way to represent this equation is by combining T f and T i into one variable T, which is the change in the temperature; the same thing as T f T i. Lets look at some examples of typical problems you will see: Final temperature minus the inititial temperature
2 Example 1: How much heat (in calories) is needed to raise 20 grams of water from 5 o C to 40 o C? m = 20 grams T f = 40 o C T i = 5 x = (20 g) (1 cal/g o C)( C) = 700 cal Example 2: How much heat (in calories) is needed to raise 140 grams of water from 20 o C to 25 o C? m = 140 grams T f = 25 o C T i = 20 o C x = (140 g) (1 cal/g o C)( ) = 700 cal Please note that the amount of calories is the same in both problems. In example 1 we have 20 grams being changed by 35 o C while in example 2 we have much more water (140 grams) being changed by a much smaller amount (5 o C). Example 3: How much heat (in calories) is needed to raise 250 grams of water from 80 o C to 87 o C? m = 250 grams T f = 87 o C T i = 80 o C x = (250 g) (1 cal/g o C)( C) = 1750 cal
3 Example 4: How many calories are needed to raise 50 grams of iron from 55 o C to 200 o C? The specific heat capacity of iron is 0.11 cal/g o C. Example 5: m = 50 grams c = 0.11 cal/g o C T f = 200 o C T i = 55 o C x = (50 g) (0.11 cal/g o C)( C) = cal How many grams of aluminum can be heated from 90 o C to 120 o C if 500 calories are applied? The specific heat of aluminum is 0.21 cal/g o C. Example 6: Q = 500 calories m = x grams c = 0.21 cal/g o C T f = 120 o C T i = 90 o C 500 cal= (x) (0.21 cal/g o C)( C) x = 79.4 grams What is the specific heat capacity of a substance if 400 calories cause 25 grams of it to go from 60 o C to 190 o C? Q = 400 calories m = 25 grams c = x cal/g o C T f = 190 o C T i = 60 o C 400 calories = (25 g) (x)(130 0 C) x = cal/g o C
4 Example 7: What is the final temperature if 500 calories are applied to 40 grams of copper at 20 o C? The specific heat capacity of copper is cal/g o C. Example 8: Q = 500 m = 40 grams c = cal/g o C T f = x T i = 20 o C 500 cal = (40 g) (0.092 cal/g o C)(x 20 0 C) x = 156 o C What was the initial temperature if 250 calories were applied to 100 grams of gold and the final temperature of the gold was 175 o C? The specific heat capacity of gold is cal/g o C. Example 9: Q = 250 calories m = 100 grams c = cal/g o C T f = 175 o C T i = x o C 250 cal= (100 g) (0.031 cal/g o C)(175 x 0 C) x = 94 o C What is the change in the temperature if 75 calories are applied to 10 grams of water? Q = 75 calories m = 10 grams c = 1 cal/g o C T f T i = x (note that in this problem we want the change so both T f T i are included as x) 75 calories = (10 g) (1 cal/g o C)(x) x = 7.5 o C
5 Questions Specific Heat Capacities Substance Specific Heat Capacity (cal/g o C) Water 1.0 Ice 0.49 Copper Gold Iron 0.11 Aluminum How many calories would it take to raise the temperature of 200 grams of water from 5 o C to 85 o C? 2. How many calories would problem number 1 be if it was aluminum instead of water? 3. How many grams of copper could be heated from 20 o C to 75 o C if 1200 calories are applied to it? 4. How many grams of iron could be heated from 15 o C to 300 o C if 8000 calories are applied to it? 5. What is the specific heat capacity of a substance if 750 calories caused 100 grams of it to go from 90 o C to 135 o C? 6. When 400 calories are applied to 20 grams of a substance, it goes from 62 o C to 82 o C. Which of the substances in the table above is this? 7. What is the final temperature if 400 calories are applied to 225 grams of iron at 40 o C? 8. What would the final temperature be if 500 calories are applied to 150 grams of ice at 90 o C? 9. What is the temperature change if 50 calories are applied to 4 grams of water? 10. What would the temperature change by if a 90 gram piece of hot iron cooled by losing 200 calories. We typically think of calories as the energy from food. However, food calories are actually Calories with a capital C. This small change is very important as a Calorie is equal to 1000 calories or 1 kilocalorie. This nutrition fact shows that an egg can give 70 Calories which would be 70,000 calories. This is enough energy to raise 70,000 grams of water by 1 o C! A typical diet is supposed to have about 2000 Calories or 2,000,000 calories which seems like a lot. Keep in mind, though, that our bodies have to keep us at about 31 o C every minute of the day and a typical person weighs around 175 pounds or around 80,000 grams. Also, our bodies are not perfect in turning every bit of that energy into useful, productive outcomes.
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