Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > MCAT Study Question Q&A > q = mcat general question PDA View Full Version : q = mcat general question Pamplemousse12312-29-2010, 01:21 PMHi there, I need some clarification with this eqn: q = mC delta T if we decrease the mass by 1/2, then the heat released or absorbed decreases by half. This suggests that the temperature change is reduced by half. Does this violate the eqn above? q/2 does not equal to m/2 C (delt T)/2 but rather q/4 . :eek: please explain! upchurch12-29-2010, 01:31 PMHi there, I need some clarification with this eqn: q = mC delta T if we decrease the mass by 1/2, then the heat released or absorbed decreases by half. This suggests that the temperature change is reduced by half. Does this violate the eqn above? q/2 does not equal to m/2 C (delt T)/2 but rather q/4 . :eek: please explain! So, specific heat capacity (C) is the amount of ENERGY required to change the temperature by 1 unit of temperature per unit mass. In this case, you are decreasing the mass by half, so you you need half the amount of energy (q) to get the SAME temperature change. So, while q and m change from case 1 to case 2, the delta T does not. Halfling12-29-2010, 01:36 PMHi there, I need some clarification with this eqn: q = mC delta T if we decrease the mass by 1/2, then the heat released or absorbed decreases by half. This suggests that the temperature change is reduced by half. Does this violate the eqn above? q/2 does not equal to m/2 C (delt T)/2 but rather q/4 . :eek: please explain! Umm.. this is a little confusing.. Can you explain what you were trying to say again? For what I understood in your question: Q = mc(delta T) If you half the mass, m, the you will half the heat. Yep! An example, if you have a metal block that is heated to say 100 degrees and then you have a smaller metal block (half the mass) heated to 100 degrees then you will effectively gain more heat back from the bigger block. This is great cause if you're in the mid-west and its real cold then you would use the bigger block to stay warm. :) * Notice the temperature is the same in both cases but if you imagine being cold and you have one metal block mass m and another metal block mass m/2.. You would stay a lot warmer with the larger mass. The temperature is not affected. If you half both mass and the change in temperature you would get a lot less heat back i.e., q/4... If you think of it it makes sense. Use the same block example discussed above. Now, compare a block of mass, m/2 and initial temperature of 50 degrees... you would get a lot less heat then m/2 and temp initial of 100 degrees... (Let's assume the final temperature in all these cases is 0 degrees.) Hope that helps... Your question was a little confusing, please explain a little more if I'm nowhere close to answering your question. Pamplemousse12312-30-2010, 06:07 PMSorry for not being clear, let me be more specific: #45 pg 175 from TBR asks: Adding 1.21g of Mg into Beaker 1 instead of 2.43g would have led to a temperature increase up to what temperature? Note, in the passage it said that the inital temp was 22degC and with 2.43g of Mg, the temperature difference was 16degC (or a final temp of 38). The answer, according to TBR, is that adding 1/2 the amount of Mg, "results in only half of the heat released that was released in expt 2. the temperature increase in beaker 2 (using 2.43g) was 16deg C, so the temp increase using 1.21 g should be about 8degC," yielding a final temp of 30degC. Now when I thought about this with the eqn: q = mCdeltT, I got confused because the numbers do not cancel out. If we decrease the mass by .5, and the heat decreases by .5, the temperature change should not change - it should remain 16degC. But, that's not the answer! This is why I need further clarification on this question. If you know what's the problem in my reasoning, please let me know! Thanks! WhiteWashed12-30-2010, 06:27 PMHi there, I need some clarification with this eqn: q = mC delta T if we decrease the mass by 1/2, then the heat released or absorbed decreases by half. This suggests that the temperature change is reduced by half. Does this violate the eqn above? q/2 does not equal to m/2 C (delt T)/2 but rather q/4 . :eek: please explain! I think i see what you are doing....if you do m/2, it will result in q/2...but delta T will be the same! If you take 50grams of water and decrease it by 50degrees...it will result in 50*50*4.2 joules of heat... If you take 25 grams out of the initial 50grams of water and do the same thing....the heat flow will be 25*50*4.2, or half of the original q. makes sense? Pamplemousse12312-30-2010, 06:39 PMI think i see what you are doing....if you do m/2, it will result in q/2...but delta T will be the same! If you take 50grams of water and decrease it by 50degrees...it will result in 50*50*4.2 joules of heat... If you take 25 grams out of the initial 50grams of water and do the same thing....the heat flow will be 25*50*4.2, or half of the original q. makes sense? Nope, TBR mentioned with two questions (one of them is above) that if you decrease the mass by .5, q is decreased by .5, and the change in temp is decreased by .5. Clearly, I'm missing out on something here! WhiteWashed12-30-2010, 06:41 PMNope, TBR mentioned with two questions (one of them is above) that if you decrease the mass by .5, q is decreased by .5, and the change in temp is decreased by .5. Clearly, I'm missing out on something here! not exactly sure why the change in temp would be half of what it is orginally.... the change in temp doesn't take into account how much or how little there is a cup of water at room temp will decrease in temp the same amount a gallon of water would. Unless im wrong??? BerkReviewTeach12-31-2010, 10:35 AMNope, TBR mentioned with two questions (one of them is above) that if you decrease the mass by .5, q is decreased by .5, and the change in temp is decreased by .5. Clearly, I'm missing out on something here! Was the mass going down by half or the concentration going down by half? What question number are you looking at, because I think there is some fact you might be overlooking in the question or passage. Pamplemousse12312-31-2010, 11:02 AMWas the mass going down by half or the concentration going down by half? What question number are you looking at, because I think there is some fact you might be overlooking in the question or passage. It's on pg 175, number 45, passage 7 of the thermochem section. answer explanation is on pg190. Let me know! Rabolisk12-31-2010, 11:29 AMI don't have the BR book, but my intuition says that you misapplied the formula. My guess is that the TEMP change they are referring to is not the temp change of the Mg, but of the entire beaker + solutions. The mass of the beaker changes very little (i.e. it is constant). Heat is released or absorbed when a substance is solvated or undergoes a chemical reaction (e.g. enthalpy of combustion, enthalpy of solvation). The amount of heat released/absorbed is dependent on the amount of substance, so halving the amount (mass) of magnesium results in HALF the heat. Now take that and apply it to the formula q = mc(delta)T with the same mass and q/2. You get T/2. What BRTeach was saying is the same thing that I'm saying, so I think my explanation is correct. Pamplemousse12312-31-2010, 11:35 AMI don't have the BR book, but my intuition says that you misapplied the formula. My guess is that the TEMP change they are referring to is not the temp change of the Mg, but of the entire beaker + solutions. The mass of the beaker changes very little (i.e. it is constant). Heat is released or absorbed when a substance is solvated or undergoes a chemical reaction (e.g. enthalpy of combustion, enthalpy of solvation). The amount of heat released/absorbed is dependent on the amount of substance, so halving the amount (mass) of magnesium results in HALF the heat. Now take that and apply it to the formula q = mc(delta)T with the same mass and q/2. You get T/2. What BRTeach was saying is the same thing that I'm saying, so I think my explanation is correct. That makes good sense! This is a good wake up call that I can't just stick anything into a formula. Thanks! :) BerkReviewTeach01-02-2011, 08:52 AMOn this question, and ones like this, you must be careful with the mass. The mass you use in that formula is the mass of the material that is heating up (the calorimeter contents). You are adding Mg to water, so Mg is the reactant while water is the material that is heating up. So when you cut the mass of Mg in half, you get half the heat (q is half). But, the mass of the surrounding water absorbing that heat has not changed, so m in the equation reamins the same. So we have q = mCdeltaT, where q is now half, m is the same, and C is the same (any change because of solute concentration is trivial and can be ignored based on the aqnswer choices). This means that deltaT must be half as much this time. Hopefully this seems clear, because it's a pretty common point in calorimetry passages and questions. It should make the next calorimetry passage easier. BerkReviewTeach01-02-2011, 08:53 AMI don't have the BR book, but my intuition says that you misapplied the formula. My guess is that the TEMP change they are referring to is not the temp change of the Mg, but of the entire beaker + solutions. The mass of the beaker changes very little (i.e. it is constant). Heat is released or absorbed when a substance is solvated or undergoes a chemical reaction (e.g. enthalpy of combustion, enthalpy of solvation). The amount of heat released/absorbed is dependent on the amount of substance, so halving the amount (mass) of magnesium results in HALF the heat. Now take that and apply it to the formula q = mc(delta)T with the same mass and q/2. You get T/2. What BRTeach was saying is the same thing that I'm saying, so I think my explanation is correct. Sorry I posted before reading your response above, because it is perfect. You explained it really well. I basically double posted. TooSerious07-03-2012, 12:47 PMSorry for not being clear, let me be more specific: #45 pg 175 from TBR asks: Adding 1.21g of Mg into Beaker 1 instead of 2.43g would have led to a temperature increase up to what temperature? Note, in the passage it said that the inital temp was 22degC and with 2.43g of Mg, the temperature difference was 16degC (or a final temp of 38). The answer, according to TBR, is that adding 1/2 the amount of Mg, "results in only half of the heat released that was released in expt 2. the temperature increase in beaker 2 (using 2.43g) was 16deg C, so the temp increase using 1.21 g should be about 8degC," yielding a final temp of 30degC. Now when I thought about this with the eqn: q = mCdeltT, I got confused because the numbers do not cancel out. If we decrease the mass by .5, and the heat decreases by .5, the temperature change should not change - it should remain 16degC. But, that's not the answer! This is why I need further clarification on this question. If you know what's the problem in my reasoning, please let me know! Thanks! I understand BerkReviewTeach and Rabolisk, but why in the world are they even mentioning Experiment 2 at all in the explanations? The question stem specifically refers to Beaker 1, which although Experiment 2 has an identical set up, Beaker 1 is local to Question 45.