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Numerade Educator

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Problem 4 Medium Difficulty

The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at $20.0^{\circ} \mathrm{C} :$ (a) water; (b) concrete; (c) steel; and (d) mercury.

Answer

(a) $21^{\circ} \mathrm{C}$
(b) $25^{\circ} \mathrm{C}$
(c) $29.3^{\circ} \mathrm{C}$
(d) $50.3^{\circ} \mathrm{C}$

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Video Transcript

okay for this question, we have one kill, a calorie of heat being transferred into one kilogram of mass for four different types of matter. On you have an original temperature t of 20 degrees Celsius, and we're looking for the final temperature of each of these. So or this problem? We want to use Thekla sick heat equation, which is cute, the heat equals and the mass times specific specific heat. The material see times Delta T so Delta T It's the same as final temperature minus initial temperature so we can rewrite this equation to say Q equals M C terms t f minus T o can divide both sides by M. C. Says gives this t f minus t o. So the equation will be working with to solve for t f is I can draw them here. Yeah, is the original temperature plus Q over M. C. So let's just highlight this. We can remember it and separate it from where we're going to solve for these. Okay, So part a of the question is asking, what is the temperate final temperature If the matter involved is water, so the specific heat for water you have to look this up and this is one calorie program, times degrees Celsius. So we plug this into our equation. You have t final equals t not which is 20 degree Celsius plus Q. And all of these is one kilocalories. That's 1000 calories divided by M times C. So em is the same for all of these. It's one kilogram or 1000 grams. We're gonna work in kill groups and our calories and grams, rather because our specific heats are given in calories and grams. That's why I'm converting from kilograms and kill cows. So cue over M. C. So we need that specific heat here, which is one calorie over grams times degrees Celsius. So we have a calorie over a calorie graham over Grandma's cancel out. We also have 1000 over 1000 such as one. So this is 20 degrees Celsius plus one degrees Celsius because this the degree Celsius comes up top. So this gives us our answer for water of 21 degrees Celsius on circle that Okay, so, for B, you have same thing. But our specific heat is so your 0.2 calories per gram degree Celsius. So our final temperature is 20 degrees Celsius seems above. Plus we know now that our calories and grams we're gonna cancel out the thousands are gonna cancel out. So this is just one over the specific E J 0.2 Celsius and that is 20 plus five degrees Celsius. Gives us 25 degrees Celsius. That's our answer for beat. See, our material is steel, says was if you keep for steel. If you looked this up, is your 0.12 calories per gram story Celsius so T final is 20 degree Celsius initial, plus cancel everything out. We know it's just one over the specific heat at this 10.0 point 123 Celsius, which is 20 plus 8.3 three Celsius. Just put this nor calculator real quick. Come, you can give me a 0.3. So that is correct. 28.3. It's a decree, okay? And the last one is mercury, which has a specific heat of what's your 0.0 33 So TF, we know now is 20 degrees Celsius plus one oversee, which is your 10.33 degrees Celsius. Put that in your calculator and its 20 plus 30 over 30.3, gives us 50.3 degrees Celsius for a final temperature. So there's our four answers for A, B, C and D for this problem.