IDS 102 Answers for Specific Heat End of Module Questions

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IDS 102 Answers for Specific Heat End of Module Questions 1. A cup with 40 grams of 90 C water is mixed with a cup with 70 grams of water at 10 C. a. Predict the final temperature of the water. Explain your reasoning. If the amount of water in each part was equal, the temperature would be about 50 C, however there is almost twice as much 10 C as 90 C water, so the temperature should be between 10 C and 50 C. Some students have figured out a clever way to predict the final temperature of two objects (water in this case) with different mass as long as they are made by the same material. Here s the deal: if the two bodies of water have the same mass, then we just take the average: you would add the TWO temperatures and divide by TWO. If you were mixing three bodies of water, all with the same mass, you could still take the average, right? You d just add THREE temperatures and divide by THREE. We can interpret this problem, 40 g on one side and 70 g on the other, as a mixture of 110 bodies of water, all of which have a mass of one gram. You just add up 40 of the temperatures at 90 degrees and 70 of the temperatures at 10 degrees, and then divide by (40 + 70) which is 110. In equations that is: 40 g 90 + 70 g 10 40 g + 70 g = 4300 g 110 g 39 (this is called a weighted average ) b. Calculate the expected final temperature of the water. (Assume that no heat is lost to the surroundings). You could solve this in a couple of different ways, but here is one method: Set-up a table similar to the ones you used in class: Calories Transferred Temperature of 40 g of hot water Temperature of 70 g of cold water 0 1600 400 40 90 o C 50 o C 40 39 10 o C 32.8 38.5 39.1 So again we see that the final temperature should be about 39 C

2. 10 calories of heat were added to both a sample of water and a sample of copper. The specific heat of copper is around 0.1 cal/(g ). a) Review question: what is the specific heat of water? 1.0 cal/(g ). b) The 10 calories of heat increased the temperature of a sample of water by 5 C. What was the mass of the water? Here ya got yer basic water, and we know that if we had one gram of the stuff then those ten calories would have increased the temperature by 10 C. Is that what happened? Nosirree. The temperature only went up by half that much. That means that it took TWO calories to raise the temperature of the sample by 1 C, which is just another way of sayin that the heat specific heat of this was 2 calories/(g C). When we raise the temperature of water by 1 C, then we know each gram received 1 calorie and since there were 2 calories bein handed out, there must have been two grams of water there to absorb em. c) If the sample of copper has the same mass of the water, will it change temperature to a greater or lesser degree than the water? We see that the specific heat of water is about TEN TIMES greater than the specific heat of copper (if your experiments told you it was 11 times greater, then congratulations, because the exact answer is somewhere between 10 times and 11 times greater). Since water has a greater specific heat: It takes MORE HEAT to make the SAME MASS of water have the SAME TEMPERATURE CHANGE as the copper. Given the SAME AMOUNT OF HEAT and the SAME MASS, water will have a SMALLER CHANGE IN TEMPERATURE than the copper. In fact if the masses were the same the copper would have a temperature change that is about TEN TIMES as great as the water (between 10 and 11 times greater).

3. A scientist is given 3 samples of an unknown material to examine. Each sample weighs ten grams. She adds 100 calories of heat to each substance and notes the following: Sample A increases temperature by 10 Sample B increases temperature by 50 Sample C increases temperature by 20 a) Which sample has the lowest heat capacity? Which has the highest? Explain your reasoning. First a little bit about vocabulary: heat capacity vs. specific heat Specific heat is the amount of heat required to make a certain temperature change PER GRAM OF MATERIAL. Substances have given specific heats depending on the material. Copper has a specific heat. Liquid water has a specific heat. Ice has a different specific heat from liquid water, but all pure ice has the same specific heat as other samples of pure ice. All pure water has the same specific heat as well. Specific heat is a property of the material Heat capacity is a property of an object, and it depends on the size of the object. It is the amount of heat required to change the temperature OF THAT OBJECT by a certain amount. If you remove a cup of water from Lake Washington, the lake has a higher heat capacity than the cup of water simply because it is so much bigger. The heat capacity is actually the specific heat multiplied by the mass of the object. This question uses the phrase heat capacity but since all of the objects have the same mass, it doesn t matter whether you think about heat capacity or specific heat in this case. In a more general setting, if two objects are made of the same material but one is larger than the other, the larger object has a greater heat capacity. Now back to the original question The masses were the same and the amount of heat that was added was the same for each of the three samples. Something with a large specific heat needs A LOT OF HEAT to produce a GIVEN TEMPERATURE CHANGE (for a given mass). Put another way, given the SAME AMOUNT OF HEAT, the thing with the highest specific heat will CHANGE TEMPERATURE THE LEAST (it needs more heat to have a temperature change as great as the other samples). So Sample B has the lowest heat capacity. Sample A had the highest heat capacity. b) If a warm sample A was mixed with a cold sample B, which sample would show the greatest temperature change? Remember all samples have the same mass. The heat lost by sample A (in calories) would equal the heat gained by sample B (in calories). Since sample A has the greatest specific heat, it will change temperature the least. Sample B would have a temperature change that is five times as great.

4. In a lab experiment, an 80 g sample of water at 20 o C is mixed with 80 g of aluminum at 50 o C. The specific heat of aluminum is 0.20 cal/g o C. In other words, it takes only 0.2 calories to raise the temperature of 1 g of aluminum by 1 o C. NOTE: THE SOLUTIONS THAT FOLLOW ARE MUCH MORE DETAILED THAN YOU NEEDED TO ANSWER A SHORTENED VERSION OF THE QUESTION. THIS COMPLETE SOLUTION HAS BEEN INCLUDED FOR YOU TO STUDY IF YOU WANT TO SEE ALL OF THE DETAILS OF THE INTERACTION. a. What is your estimation of what the final temperature of the mixture will be? (Do not do any calculations yet). Explain your reasoning. Well, the specific heat of aluminum is only about one-fifth of the specific heat of water. That means that 80 g of aluminum has the same specific heat as 16 g of water. In this case we expect the final temperature to be a lot closer to the initial temperature of the 80 g of water, so I m guessing it will be around 25 C, or maybe 30 C. b. How many calories are required to change the temperature of 1 gram of aluminum by 5 o C? Show your work. Just one calorie (not Calorie) Each gram requires only 0.2 calories to increase in temperature by 1 C, so 1 g will require five times as much heat, or 1.0 calories to increase in temperature by 5 C. c. How many calories are required to change the temperature of 80 grams of aluminum by 5 o C? Show your work. For 80 grams of aluminum we will need 80 times as much heat, or 80 calories. d. Complete the following table by filling in the number of calories needed to change the temperature of the aluminum by 5 o C, and the resulting temperature change of the cold water. Continue until you have determined the final temperature of the mixture. (Note: if you have trouble doing this in 5 o C increments, you can do it in 1 o C increments, but it will take many more steps) Total Calories Transferred 0 80 160 240 320 400 Temperature of 80 g of aluminum 50 o C 45 o C 40 C 35 C 30 C 25 C Temperature of 80 g of cold water 20 o C 21 C 22 C 23 C 24 C 25 C e. What is the final temperature of the mixture? 25 C f. How many total calories are transferred? 400 calories.

g. What is the temperature change of the water in this experiment? 5 C h. What is the temperature change of the metal in this experiment? 25 C i. Compare the heat lost by the metal to the heat gained by the water. Explain your reasoning. The heat lost by the piping hot metal is EXACTLY THE SAME as the heat gained by the oh-so-cool water. All of the heat lost by the metal goes into the water. Honest.