hypebeast Posted June 13, 2014 Report Share Posted June 13, 2014 solve this 7 kg of water at 30 °C add a 3 kg ball of lead to it at 100 °C waters specific heat: 1000 J/KgC lead specific heat: 130 J/KgC calculate the average temperature or w/e they call it in your country. Link to comment Share on other sites More sharing options...
Enrju Posted June 13, 2014 Report Share Posted June 13, 2014 Do you have an answer? Link to comment Share on other sites More sharing options...
Enrju Posted June 13, 2014 Report Share Posted June 13, 2014 So my answer is 33,69°C, which is probably wrong because i just finished 10th grade. Link to comment Share on other sites More sharing options...
Thar Posted June 13, 2014 Report Share Posted June 13, 2014 Gosh, it's been so long since my last Physics class... and my teacher sucked at his job. All I know about this problem is that there's a lot of algebra involved. I'll see what I can do. Link to comment Share on other sites More sharing options...
Βyakuya Posted June 13, 2014 Report Share Posted June 13, 2014 I don't take physics, but I did take AP Chemistry not to long ago, and luckily Thermodynamics is one thing I'm glad to see and scared to see again. But following the laws: q=mcΔt Which is the internal heat of an object (q) equal to mass (m) times specific heat (c) and change in temperature (Δt) Then -q=mcΔt (Of Lead Ball) = q=mcΔt (Of Water), because heat lost from the lead ball flows to the water. Some heat may flow to the container but we disregard that; 1rst Law of Thermodynamics FTW. So... -(7kg)(130 J/KgC)Δt = (3kg)(1000 J/KgC)Δt Well I thought water had a normal specific heat of 4.18 J/KC, but whatever. If I knew the Δt then I could solve it but this by sending one side to the other and computing it algebraically. I guess the average change in temp would be the changer in temperature as we call it.This should inspire you......maybe. Link to comment Share on other sites More sharing options...
Aix Posted June 13, 2014 Report Share Posted June 13, 2014 Water's specific heat capacity is 4181 J/KgC though. Anyway, I'll solve this without subbing in the numbers until the very end to create a simple final derived formula that can be used for anything and very quickly. That's just the way I like to do things. M1=Mass of Water M2=Mass of Lead T1=Temperature of Water T2=Temperature of Lead C1=Specific Heat of Water C2=Specific Heat of Lead F=Final Temperature Q1=Thermal Energy of Water Q2=Thermal Energy of Lead Qgain+Qloss=0 M1*(F-T1)*C1+M2*(F-T2)*C2=0 M1*F*C1-M1*T1*C1+M2*F*C2-M2*T2*C2=0 M1*F*C1+M2*F*C2=M1*T1*C1+M2*T2*C2 F*(M1*C1+M2*C2)=M1*T1*C1+M2*T2*C2 F=(M1*T1*C1+M2*T2*C2)/(M1*C1+M2*C2) or F=(Q1+Q2)/(M1*C1+M2*C2) Holy does that look confusing as fuck when typed. Anyway, for subbing in the numbers now, I'll use the specific heat that you've given. F=(7*30*1000+3*100*130)/(7*1000+3*130) F=(210000+39000)/(7000+390) F=249000/7390 F=33.6941813261 Link to comment Share on other sites More sharing options...
Smear Posted June 14, 2014 Report Share Posted June 14, 2014 um doesnt physics have like s=d/tbc i remember thatand thats about it.I did Physics for all of last year but I didn't listen in class at all, and this year I dropped down to Senior Science. which is pretty much just and easy version of all 3 of the basic sciences. (Physics, Chemistry, Biology) Link to comment Share on other sites More sharing options...
Trebuchet MS Posted June 15, 2014 Report Share Posted June 15, 2014 First, let's assume a closed system. This means no heat is gained or loss due to the environment. Next, let's assume his given specific heat capacities are true, because specific heat capacity of water is commonly rounded to 4200 J/kgC. If we put it in terms of calories, then that would be 1000 cal/kgC. Total internal energy of water: 7 x 30 x 1000 = 210,000 J = 210 kJ Total internal energy of steel ball: 3 x 100 x 130 = 39,000 J = 39 kJ Total internal energy of system = 249 kJ When the system is at thermal equilibrium, both objects are at the same temperature. Let's call this temperature T. Because we have a closed system, the total internal energy will remain the same. Hence... 7 x T x 1000 + 3 x T x 130 = 249,000 7000T + 390T = 249,000 7390T = 249,000 T = 33.694 C correct to 3 significant figures. Link to comment Share on other sites More sharing options...
'tyleR Posted June 23, 2014 Report Share Posted June 23, 2014 So my answer is 33,69°C, which is probably wrong because i just finished 10th grade. 2 posts and probably only 15 years old or some shit and he's already better than this whole muthafucking site. Good job, kid. Link to comment Share on other sites More sharing options...
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