What is the heat capacity of a metal slug if a temperature rise from 25°C to 54°C requires 135 J?
963644
Heat capacity = q/(delta)T\n\nC= 135J/(54°C-25°C)\nC= 135J/25°C\nC= 5.4 J/°C
574253
1 slug = 8.78865485 × 10^27 atomic mass units\nthus\n2.8283 slugs = 2.48569525 × 10^28 atomic mass units
750705
212 F, 100 C,     373 K
954796
51kg x 25 m = energy used\n\ndivide the energy used by 0.30 to get the second answer
279992
Copper's resistance increases by a factor of 0.0068 per degree C (ref.). This resistance increase is linear, not exponential. The formula is Resistance = R(initial) * [1 + alpha * (T(final) - T(initial)], where alpha is the temperature coefficient. Also note that self-heating due to any current flow will add to the ambient temperature effect, so power dissipation in a cable will be higher than what is calculated from resistance based on ambient temperature alone.
807123
Your reaction evolved 500 calories (you raised 100 grams of water five degrees) , or 500*4.184 J = 2092 Joules.\n= 2.092 kJ\n\n\n2.092 kJ / 0.05225 mol = 40.0 kJ/mol
1161541
Ha ha :))\n\n\nlet's say x be the length of metal rod 1 in the final composite rod \n\n===> (31-x) is the length of matal 2 in composite rod \n\nalpha 1(coefficient of linear expansion for metal 1 )  = change in length /original length  *change in temp =6.35* 10.^-2/31*100\n=\n\nSimilary ,for metal 2\n\n\nalpha 2  = (3.20 * 10.^-2)/31*100\n\nFOR THE FINAL compsite  rod  \n\nx*alpha1 * change in temperature + (31-x)* alpha 2* change in temp = 5.55 * 10.^-2 \n\nchange in temp above eqn is same =100\n\n0.002x+(31-x)*0.0010=5.55*10.^-2\n\n0.001x=0.0245\n\nx= 24.500 cm of metal rod 1\n\n& (31-24.500) cm = 6.500 cm of metal rod 2\n\n\nTHE author below has made a mistake of leaving the orriginal length :\n\nThe eqn is:\n\ndl=alpha*original length*dT
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
820184
I think you should break things down and think of the question like this.\n\nThe heat for a process is 250.3J, Its doubled with the same temperrature difference, the temperature difference is 5C.
266721
The heat source was hot enough to make the rise of the temperature from 32 to 145 F. The rise is 113 F obviously .\nIt took 4 x 60 minutes. So the average increase of the temperature is\n(113 F)/(240 minutes) = 0.47 F/min.
904003
Temperature rise of water = 2 deg C = 275 K\nq=mc(change in temperature)\n  = (100)(4.18)(275)\n  =114950J\n  = 114.95 kJ\nHmm. Either I'm wrong or the question is incomplete. Did the question perhaps give more information on the reaction?
562654
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
977787
Are you asking about thermal conductivity or specific heat?\nAnd why are you not doing the research you should be doing:\nhttp://www.engineeringtoolbox.com/specific-heat-metals-d_152.html\nhttp://www.engineeringtoolbox.com/thermal-conductivity-metals-d_858.html
858350
4.184 J = 1 calorie\n\n4.184 * 542 =   x\n  x= 2267.7 Joules
705010
The specific heat of water is 4.184 J/g C.\n\nSince water's density is 1 g/ mL, 500 mL of water will have a mass of 500 g.\n\nThe equation to calculate the joules of energy required is:\nheat = mass x specific heat x change in temperature\n\nenergy = 500 (4.184)(95-25)\nenergy = 146440 J or 146.44 kJ\n\nHope that helped.
955628
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.