if 200g of water at 20 degrees celcius absorbs 41840J of heat, what will its final temperature be?
366921
Q = mc delta T\nHeat = mass times specific heat times change in temperature\nspecific heat for water is 4.186 kJ/kg C\n-41840 J (negative because absorbed) = .200 kg x 4.186 x 10^3 J/ kg C x (Tf - 20)\nTf = final Temp.\nsolving the equation gives you Tf = -30.0 degrees Celsius
422010
Its easy...\nHeat lost by meat = heat gain by contents in calorimeter(ie water)\n\nm X s X delta(t) = M X SX Delta(T)\nm-mass of lead                  M-mass of water\ns-specific heat of lead         S-Specific heat of Water\nt=change in temp=(373-x)              t-Change in Temp = (303-x) because you dont know final temerature\ns of lead= 0.032cal/g (i think)  S of water= 1cal/g\n\nsubstitute these values .. you will  get 53 degrees C as final answer\n\nI hope you got the point.\nGood luck\n\nNote: there may be some calculation mistakes, but the theory is the same
1064976
Q (heat energy) = m(mass in gram)*c(specific heat in J/gm/K)*t (change in temp)\nQ = 25 * 1 * (36-24)\nQ = 300 J
1081570
final temperature is 71 degrees celsius
715037
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
470685
You must calculate difference of enthalpy between first and second condition.  Enthalpy depends on  temperature and pressure.And it can be find from thermodynamic tables.\n\nI accept 1 bar for water's pressure \n\n1 bar and 20 degree water's enthalpy = 84 kj/kg \n1 bar and 55 degree water's enthalpy = 104,725 kj/kg \n\ndifferent between inlet enthalpy and outlet enthalpy = 104,725-84 = 20 kj / kg (this heat necessary to increase of 1 kg water's degree from 20 C to 55 C ) \n\namount of water = 6000 lt = 6 m3\nspecific density of water which is at 20 C and 1 bar = 0,0010017 m3/kg\nmass of water = 6 / 0,0010017 = 5990 kg \n\ntotal heat requirement = 5990 kg x 20 kj/kg = 119800 kj \n\nXXX if this proses time is 1 hour then \n\npower or capacity of boiler for perfect situation ( perfect situation = no losses and boiler effiency 100 %) = 119800 / 3600 second = 33,2 kw \nif boiler effiency = 85 % then \nreal capacity of boiler for proces time is 1 hour = 33,2 x (100/ 85) = 39,05 kw \n\nXXX if this proses time is a half hour then \n\npower or capacity of boiler for perfect situation ( perfect situation = no losses and boiler effiency 100 %) = 119800 / 1800 second = 66,55 kw \nif boiler effiency = 85 % then \nreal capacity of boiler for proces time is 1 hour = 66,55 x (100/ 85) = 78,3 kw \n\nif you want to calculate capacity of boiler ,  first you must know proses time and pressure.
218320
Q=m*c*(delta T)\nWhere Q is heat added, \nm is mass, \nc is specific heat of substance (for water=4.186 joule/gram °C)and\ndelta T is change in temperature. \n\nSo delta T = Q/(m*c)= 31500/(750*4.18)=9402.98 degrees Centigrade
1119466
I donno ;=)\n\nMay be start with a heat balance equation like\n\nQfinal=Qwater-start  - Qice-start – Qlatent\n\nIn other words the final heat of the system is equal to the original amount of heat minus the heat required to bring ice to the melting point and minus again the amount of heat needed to melt the ice\n\nOr \nQfinal=(Specific-heat of water) (temp in K)(mass-final)\n\nmass-final= mass of water + mass of ice= m1 + m2\n\nNote -273.16 °C =0 degrees on Celsius scale = 273.16 °K (273 I hope is good enough for you?)\n\nQwater-start  =( 4190 J/(kg*K) (t-start K degrees )(m1)( 4190 J/(kg*K)=A(m1)  A= 4190 (76.2 + 273)\n\nQice-start= 2100 J/(kg*K)( - 23.6+273)(m2)= B(m2)\n\n\nQlatent= 334 kJ/kg(m2) =C(m2)\n\nWe have \n\nD(m1+m2)=A(m1)+B(m2)+C(m2)  you know that m1=.280kg  then you figure out m2? \nAnd \nA= 4190 (76.2 + 273)\nB=2100 (- 23.6+273)\nC=334,000\n\nActually Specific heat of ice = 2060 J/(kg*K) not 2100 J/(kg*K)\nAnd Specific heat of water  =4,183  J/(kg*K)
237697
O.k., so you are given the final temperature (-0.430 C). Since the freezing point of water is usually O degrees C, the temperature decrease is 0.43 degrees C. Like in the other problem, let's set up some dimensional analysis.\n\n(0.43 deg.C decr)*(1 mole compound / 1.86 deg.C decr) = X\n\nHere X would equal 0.23 moles compound\n\nNext, you want to find the grams of solute in solution.\n\n(8.02g solute/0.861 kg solvent) = Y\n\nY here is equal to 9.31 g compound/kg solution\n\nNow that you have the mass in grams and the number of moles of the solute, you can divide Y by X, which gives the units for molecular mass (g/ mol.)\n\nThe answer should be around 40.48 g/mol. The solute cannot be determined from this information because no additional information is given concerning what is most likely a compound (not a pure element).
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
845517
One way is to use the Clausius-Clayperon Equation.  See the link below and use the boiling point of water for one of the pressure and temperature pairs.  \n\nP.S.  It's equation 22 in the linked document.
177044
use the fmla.... Q=msT\n\nwhere Q is the quantity of heat required, m is mass, s is specific heat capacity of water, T is the temperature difference. i dont hav a calculator with me now, u do the computations plz...
863200
Use the equation Q=mc(delta)T.\n36.8 = .006*130*(Tf - 32)\nSolve the equation for final Temperature and you get 79 degrees Celsius
765294
First the specific heat of water is not needed because both items mixed are water and would have the same specific heat so just use the weight.\n\nso you have 60 * 80 + 40 * 25 = 100 * x    where x is the new temperature in degrees C\n\nIf the items being mixed were different then use the specific heat of each item to calculate the number of joules in each item which will give the total number of joules in the combined item.
75209
I'll answer it, but you'd best pick me for the best answer if I'm doing all this work. ;)\n\nConvert C to K:\n\n178.0C = 451.15K\n25C = 298.15K\n30.6 = 303.75K\n\nTotal energy of water initially: Ei\n\nEi = 185.0g * 298.15K * 4.184J/g*K = 2.308E5 J\n\nTotal energy of water, final: Ef\n\nEf = 185.0g * 303.75g * 4.184J/g*K = 2.315E5 J\n\nEf - Ei = 700J\n\nLet Q be the specific heat of the alloy\n\nQ * 36.0g * 451.15K = Q * 36.0g * 303.75K + 700J\n\nQ*{(36.0g * 451.15K)-(36.0g * 303.75K)} = 700J\nQ = 700J/{(36.0g * 451.15K)-(36.0g * 303.75K)} \nQ = 700J/(16241g*K - 10935g*K)\nQ = 700J/3506g*K\nQ = 0.200 J/g*K
874229
Heat released = 480*2.34 J\n\nSo 0.02 mol releases this much \n\n1 mol will release 480*2.34/0.02 = 56.16 kJ
905957
20 celsius is 293.15 kelvin