A 15.0 g sample of qu water at 30.0 °C and Assuming no heat is a answer has the correc

i have the final answer which is 19.3 degree Celsius but i dunno how to get it., i used the formula of Qloss+Qgain=0;[m*c(delta T)] + [m*c(delta T)] = 0. but my final answer is not 19.3.., can you please help me? what formula will i need to use? do i need to consider the latent heat of vaporization and fusion?if so, what should be the formula? this is my solution; Qloss(steam) + Qgain (ice) = 0 m*C (ΔT) + m*C (ΔT) = 0 160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0 76.8 (t-100) + 500 (t-0) = 0 76.8 t - 7680 + 500 t = 0 576.8 t / 576.8 = 7680/576.8 t = 13.3 °C Answers and Replies Related Introductory Physics Homework Help News on Phys.org More than half of Hudson River tidal marshes were created accidentally by humans Identifying where to reforest after wildfire Plants can be larks or night owls just like us Aug 20, 2012 #2 voko 6,054 391 maanneaguila said: do i need to consider the latent heat of vaporization and fusion? What will happen to ice exactly at freezing if you add heat to it? What will happen to steam exactly at boiling if you remove heat from it? Aug 20, 2012 #3 maanneaguila 14 0 the ice will melt and the steam will evaporate? Aug 21, 2012 #4 Curious3141 Homework Helper 2,843 87 What are the distinct processes going on here? Think steam condensation, ice melting and heating of water. Then take it in stages. First figure out how much heat is generated if all the steam condenses to water at 100 deg C. Then figure out how much heat is absorbed if all the ice melts to water at 0 deg C. There will be an "excess" of heat here, which you should compute and use later. Now you have some water (how much?) at 100 deg C and some water (how much?) at 0 deg C. Figure out the equilibrium temperature from simply mixing these, which can be done by simple proportion. Finally, use the heat "excess" you got in the first part to work out how many kelvin (or deg C) the water will go up by, add that to the previous equilibrium temperature you derived after the "mixing" step, and that's your final equilibrium temperature. Aug 21, 2012 #5 NascentOxygen Staff Emeritus Science Advisor 9,244 1,072 maanneaguila said: this is my solution; Qloss(steam) + Qgain (ice) = 0 m*C (ΔT) + m*C (ΔT) = 0 160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0 76.8 (t-100) + 500 (t-0) = 0 76.8 t - 7680 + 500 t = 0 576.8 t / 576.8 = 7680/576.8 t = 13.3 °C You won't reach a solution until you involve all 3 specific heats: that of steam, of ice, and of water. You have involved only two. The steps Curious3141 detailed should be followed. Aug 21, 2012 #6 maanneaguila 14 0 "how much heat is generated" you mean the specific heat of a substance? Aug 21, 2012 #7 NascentOxygen Staff Emeritus Science Advisor 9,244 1,072 maanneaguila said: "how much heat is generated" you mean the specific heat of a substance? A better word is "liberated". How much excess heat is liberated when that steam undergoes a phase change? Aug 21, 2012 #8 maanneaguila 14 0 where will i place the specific heat of water? since no masses are given for water. what should be the equation? :((( Aug 21, 2012 #9 NascentOxygen Staff Emeritus Science Advisor 9,244 1,072 You're not told how much water there is, no. And it's clear there is none to start with. But if you think about the problem, you'll see where the water is going to come from... NO! The steam does not "evaporate". The two substances are mixed together in a sealed flask. No mass and no heat is lost. Aug 21, 2012 #10 maanneaguila 14 0 still i didn't get it., :((( it's really hard to understand.., i really dunno what formula will i use.., i tried this formula, but still i get the wrong answer., dunno if what's the right value to substitute.. i dunno where to get or which one will i derive to get the right formula.., :'((( *sigh* Qsteam =m*C (ΔT) + mL + m*C (ΔT) = Qice = -[mL + m*C (ΔT)] Aug 21, 2012 #11 NascentOxygen Staff Emeritus Science Advisor 9,244 1,072 When equilibrium is reached, you are going to end up with your flask containing a mass M of water at temperature T. It's time to take a step back, and start over. I suggest that you set yourself the task of writing a descriptive paragraph explaining in precise detail how the container contents change from what you start with, to what you end up with. In not less than 300 words. By forcing yourself to analyze the situation step by step with sufficient clarity that you are able to make a cogent written description, you will in the process map out the method for solving it mathematically. Forget equations at this stage, if you haven't grasped what's going on, how can you be confident that you'll come up with the right equations? Aug 21, 2012 #12 maanneaguila 14 0 i got it! ) first, steam to water ; Q=m*Lv : Q= 160(540) Q=86,400 (dunno the unit) lol. ) second, ice to water ; Q=m*Lf : Q=1000(79.6) Q= 79,600 the excess heat is; (Qsteam) 86,400 - (Qice) 79,600 = 8, 600 then.., the initial temp of both subs.; Qloss(steam) + Qgain (ice) = 0 m*C (ΔT) + m*C (ΔT) = 0 160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0 76.8 (t-100) + 500 (t-0) = 0 76.8 t - 7680 + 500 t = 0 576.8 t / 576.8 = 7680/576.8 t = 13.3 °C then, the final temp of both subs. using this formula; Q = m*c (T.final - T.initial) where Q= excess heat m=mass of steam + mass of ice c=specific heat of water T.initial=13.3 Q=1160 (1) (T - 13.3) 6800 =1160T - 15428 6800+15428 = 1160T 22228/1160 = 1160T/1160 19.16 or 19.2 = T ) but still not the exact answer.., why? heheh Aug 21, 2012 #13 voko 6,054 391 maanneaguila said: then.., the initial temp of both subs.; What "both subs" do you have here? Aug 21, 2012 #14 maanneaguila 14 0 steam and ice? Aug 21, 2012 #15 Curious3141 Homework Helper 2,843 87 A few problems. You seem to be using calorie as the heat unit, which is acceptable, although SI units like J (Joule) are preferred. But this depends on what's covered in your course. maanneaguila said: i got it! ) first, steam to water ; Q=m*Lv : Q= 160(540)

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