Heat Transfer Calculations
Chemists measure the heat given off or taken in during a chemical reaction to determine the energy of a specific chemical or physical reaction. In this lesson, you will learn to calculate the amount of heat transferred during a physical or chemical change. Scientists use a device called a calorimeter to measure the transfer of heat during a physical or chemical change. You will be using a device like this when you conduct the Calorimetry Laboratory. It is essential that you understand the calculations required in that activity, so that you can benefit from the learning opportunity that the laboratory makes available to you. Formula for heat transfer calculations is: q = m(DT)Cp
Where q = heat transferred, DT = the change in temperature and Cp = the specific heat for the material.
The SI units for heat transferred are joules, however calories are still often used for problems involving water. You should memorize the conversion factor; 4.18 J = 1 cal. The units for specific heat are joules/grams x degrees Celsius (J/g x °C) or Calories/grams x degrees Celsius (cal/g x °C). Temperature is usually given in degrees Celsius.
Solve these problems logically and algebraically. Logically, meaning you will strive to understand the logic of performing each step, and that you will check to make sure that your answer makes sense. As in any algebra problem, you will only have one unknown. The rest of the information will be provided for you. One example of each type problem is written out below.
Type 1. Heat Transferred (q) is the unknown:
Ex. Aluminum has a specific heat of 0.902 J/g x oC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a temperature of 415.0 °C to a temperature of 22.0 °C?
Step 1: First read the question and try to understand what they are asking you. Can you picture a piece of aluminum foil that is taken out of an oven. Imagine the aluminum losing heat to its surroundings until the temperature goes from 415.0 °C to 22.0 °C.
Step 2: Write the original formula. q = m(T)Cp
Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.
q = ?
m = 23.984 g
DT = (415.0 °C - 22.0 °C) = 393.0 °C
Cp = 0.902 J/g x °C
Step 4. Substitute your values into the formula: q = 23.984 g x 393.0 °C x 0.902 J/g x °C
Step 5. Cross out units where possible, and solve for unknown. q = 8501.992224 J
Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.
q = 8.50 x 103 J Our answer makes sense because joules (J) are acceptable units for q, and the value should be positive based on the wording of the question.
Type 2. mass (m) is the unknown:
Ex. The temperature of a sample of water increases by 69.5 oC when 24 500 J are applied. The specific heat of liquid water is 4.18 J/g x °C. What is the mass of the sample of water?
Step 1: First read the question and try to understand what they are asking you. Energy is being used to change the temperature of a sample of water by 69.5 °C. What size sample of water would require 24 500 J to make that change?
Step 2: Write the original formula, and then modify it isolate the unknown.
Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.
q = 24 500 J
m = ?
DT = 69.5 °C
Cp = 4.18 J/g x oC
Step 4. Substitute your values into the formula.
Step 5. Cross out units where possible, and solve for unknown. m = 24 500 J/69.5 °C x 4.18 J/g x °C
m = 84.3344463184 g
Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.
m = 84.3 g Answer makes sense because grams are the correct units for mass, and the value should be positive.
Type 3. change in temperature (DT) is the unknown:
Ex. 850 calories of heat are applied to a 250 g sample of liquid water with an initial temperature of 13.0 °C. Find a) the change in temperature and b) the final temperature. (remember, the specific heat of liquid water, in calories, is 1.00 cal/g x °C.)
Step 1: First read the question and try to understand what they are asking you. Here they are heating up a sample of water. They want to know how many degrees increase will result from 850 calories of heat. Further, they want to know the final temperature of the water, which will simply be equal to the initial temperature + the change in temperature.
Step 2: Write the original formula, and then modify it isolate the unknown.
Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.
q = 850 cal
m = 250 g
DT = ?
Cp = 1.00 cal/g x °C
Step 4. Substitute your values into the formula DT = 850 cal/250 g x 1.00 cal/g x °C
Step 5. Cross out units where possible, and solve for unknown.
Answer to step a) DT = 3.4 °C Answer to step b) final temperature = 13.0 °C + 3.4 °C = 16.4 °C
Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.
Answers are already rounded correctly. They make sense because they show the correct units for temperature and because the final temperature is higher than the initial temperature, as it should be.
Type 4. Specific Heat (Cp) is the unknown:
Ex. When 34 700 J of heat are applied to a 350 g sample of an unknown material the temperature rises from 22.0 °C to 173.0 °C. What must be the specific heat of this material?
Step 1: First read the question and try to understand what they are asking you. Specific heat is a concept that some students struggle with. The question is about finding the specific heat by seeing how much the temperature changes when a certain amount of heat is applied. Metal heats up faster than water because it has a low specific heat. If a material has a low specific heat, the temperature change will be greater for a given amount of heat, when all other things are equal.
Step 2: Write the original formula, and then modify it to isolate the unknown.
Step 3: List the known and unknown factors. Looking at the units in the word problem will help you determine which is which.
q = 34 700 J
m = 350 g
DT = (173.0°C - 22.0°C) = 151.0 °C
Cp = ?
Step 4. Substitute your values into the formula
Step 5. Cross out units where possible, and solve for unknown.
Cp = 0.65657521286 J/g x °C
Step 6. Round to the correct number of significant digits and check to see that you answer makes sense.
Cp = 0.66 J/g x °C
Example problems: attach calculation pages showing answers boxed.
1. Determine the specific heat of a certain metal if a 450 gram sample of it loses 34 500 Joules of heat as its temperature drops by 97oC.
2. 4786 Joules of heat are transferred to a 89.0 gram sample of an unknown material, with an initial temperature of 23.0oC. What is the specific heat of the material if the final temperature is 89.5oC?
3. The temperature of a 55 gram sample of a certain metal drops by 113oC as it loses 3500 Joules of heat. What is the specific heat of the metal?
4. How many degrees would the temperature of a 450 g ingot of iron increase if 7600 J of energy are applied to it? (The specific heat of iron is0.4494 J/g xoC)
5. A 250 g sample of water with an initial temperature of 98.8oC loses 7500 joules of heat. What is the final temperature of the water? (Remember, final temp = initial temp - change in temp)
6. Copper has a specific heat of 0.38452 J/g xoC. How much change in temperature would the addition of 35 000 Joules of heat have on a 538.0 gram sample of copper?
7. How many grams of water would require 2.20 x 104calories of heat to raise its temperature from 34.0oC to 100.0oC? (Remember the specific heat of water is 1.00 cal/g xoC)
8. 8750 J of heat are applied to a piece of aluminum, causing a 56oC increase in its temperature. The specific heat of aluminum is 0.9025 J/g xoC. What is the mass of the aluminum?
9. Find the mass of a sample of water if its temperature dropped 24.8oC when it lost 870 J of heat. (remember that 4.18 J = 1.00 cal)
10. How many calories of heat are required to raise the temperature of 550 g of water from 12.0oC to 18.0oC? (remember the specific heat of water is 1.00 cal/g xoC)
11. How much heat is lost when a 640 g piece of copper cools from 375oC, to 26oC? (The specific heat of copper is 0.38452 J/g xoC)
12. The specific heat of iron is 0.4494 J/g xoC. How much heat is transferred when a 24.7 kg iron ingot is cooled from 880oC to 13oC?
Answers 1) 0.79 J/g xoC 2) 0.809J/g xoC 3) 0.56 J/g xoC
4) 38oC 5) 92oC 6) 170oC
7) 333 g 8) 170 g 9) 8.4 g
10) 3300 cal 11) 86 000 J 12) 9 600 000 J