If I want to raise 1 gram of water 2°C, how much heat should I add?
925329
You must approach this question with an understanding of energy conservation. Change in Energy = mass*heat capacity*change in temperature, where heat capacity = 1.00 for pure water. By evaluating your conditions with the above definition of thermal energy, it will show that you must supply 2 calories of heat to the system.\n\nIf you were to add another substance to the original system. You would once again approach the problem using energy conservation. The energy of a hot liquid (mass1*heat capacity*change in temperature) + the energy of a cold liquid (mass2*heat capacity*change in temperature) = mass(1+2)*heat capacity*change in temperature.\n\n-Christopher
108765
hi there,\nits like, you need to know the temp and the mass flow rate of the bottom product of the distillation column. also you need to know the boiling point of the bottom product. the mass flow rate depends on your design calculation. temp of the bottom product you should already know. and boiling point of the bottom product is readily availble.\n\nnow use m.cp.dt to find heat required to raise the temp of the product to its boiling point. where cp is the specific heat capacity of the bottom product and is available in handbooks or sometimes expressed as functions of temperature itself. and dt stands for differnce of boiling point of bottom product and the current temp of the bottom product.call this value as h1 (say).\n\nnow you know the latent heat of vapourisation of the bottom product again refer to data handbooks for the values. multiply this by the mass flow rate. and this is the heat required to vapourise the bottoms. lets call this h2 (say)\n\nnow h1+h2, gives you the heat load for the reboiler. use this heat load to calculate the steam requirement etc...\nyou can mail me sanjaykumar_16@yahoo.com for any clarifications.
1043394
Of course you must specify whether you want the thermodynamic properties of water (or chewing gum?).  Good luck.
961772
1. I'll leave it to you...\n2. Hmm.. Add some marmer rocks to make the heat spread equally in the solution.. \n\npeace\nvixklen
975732
Substance: Water\nLatent Heat Melting J/g:  335\nMelting Temp°C: 0\nLatent Heat Boiling J/g:  2272\nBoiling Temp °C: 100 \nHeat of vaporization (kJ/mol): 40.65
75392
You need to check your units.  The specific heat of water is not 4.184 J/g; it is 4.184 J/g*degrees C (or degrees kelvin).  And the heat of solution should not be in J/degree, but in J/g.\n\nTry this equation instead:\n\n[HoS = J/g] = [(Specific Heat = 4.184 J/g*C)*(mass = 100g)*(Temperature change = Initial C - Final C)] / [Mass of CaCl2 = 3 g]\n\nBy subtracting the final from initial you should get the negative value and not need the -1 at the front.\n\nSometimes enthalpy of solution is in J/mol rather than J/g, in which case you modify the above by calculating the mols of CaCl2 in 3 grams and switching that out with the 3g in the above equation:\n\n(3g) / (110.99 g/mol) = 0.027 mol\n\nTheoretically, you should want to know the heat capacity of the CaCl2 because it is also being heated, but this is assumed to be negligible.  Write that into your error analysis and you teacher will love you.  Also, make sure to check your significant figures; i don't know how many sig figs you used in weighing out your 3 grams, but that will be necessary to know how much to round everything.  Good luck.
797125
The heat lost by the warmer water will equal the heat gained by the cooler water.  The change in the heat will equal \nq = (s)(m)(T1 - T2)   where q is the heat change (units of Joules (J)), s is the specific heat of water (4.184 J/gK), m is the mass of the water in grams, and T1 and T2 are the initial and final temperatures of the water samples in Kelvin.  Assume the density of water = 1.00g/mL.\n\n(4.184J/gK)(30g)(Tf - 280K)  =  (4.184J/gK)(50g)(330K - Tf)\nTf is the final temperature of the combined solutions.\n\nSolving for Tf:\nTf = [50(330) + 30(280)] / 80  = 311K
963007
multiply the mass of the water by the specific heat and the (temperature required minus the ambient temperature).that will give you the energy in joules.if you know in how much time you want this temperature to be achieved you can calculate the power as power is nothing but the rate at which the energy is dispensed
848677
I am a HVAC trainer in the USA, and as such I have to explain how to calculate what amounts of energy it takes to do a certain heating jobs.  Looking at your question, I am of the impression that you are asking how much energy it takes to bring the temperature of 100 gallons of water from 74deg F to 90deg F.  I will explain  how to do this calculation, plus ask and answer a question that I feel might be more appropriate to an application. \n\nMost HVAC technicians define a BTU as the amount of heat energy it takes to raise one pound of water one degree Fahrenheit.  Tables will state that there are approximately 8.34 pounds of water in one gallon of water.  Taking this information, we can calculate that there are 834 pounds of water in 100 gallons of water (100 * 8.34 = 834).   So, if we raised 100 gallons of water 1deg F, we will have put 834 BTU’s into that water.  The temperature difference you talk about is equal to the difference between 74deg and 90deg which is 16deg F (90 – 74 = 16).  If one degree requires 834 BTU’s, then 16 degrees will require 16 times that amount, or 13,344 BTU’s (834 * 16 = 13,344).\n\nNow for the real issue, if you have a flow of water entering the slab at 90deg F, and it comes out of the slab at 76deg F, and if the flow rate is …… lets say 8 gallons per minute, than how much energy is the water putting into the slab?  To answer this, we use the same concept as above except we figure the flow rate in pounds per hour and multiply that times the temperature difference.  The process goes like this: (8 gallons per minute) * (8.34 pounds per gallon) * (60 minutes per hour) = 4003.2 pounds per hour.  Now multiply the pounds per hour flow rate times the temperature difference, and we get the energy going into the slab (4003.2 * 16 = 64051.2 BTU’s per hour).  So, the heat pump would have to have a capacity of a little over 5 tons (1 ton is 12,000 BTU per hour) to meet that load. \n\nNow, the energy one puts into a slab should be based on the heat loss of the structure.  It might be that you only have to put 18,000 BTU’s into the slab to offset the heat loss.  The flow rate and temperature difference can be adjusted based on the heat loss.
170755
Specific heat 0.13 J/gK
537035
Formula:  J = K*m*(T2-T1)\n\nSince both K & m are 1 for your problem, J = 125 Joules\n\nK is the specific heat defined as 1 for water.  Varies with other materials.  Also this formula is not strictly true for 1 liter unless you assume the 1 liter weighs 1 kg.  Liters are volume and the formula is for mass.
1165181
It's been awhile since i've done this but i'll give it my best\n\nFirst off, the first energy you need to calculate is the energy needed to raise the temperature of ice to 0 degrees C. so...\n\n(12.37 g)(.48 cal/deg/g)(35 deg) = +207.816 cal\n\nWhen the ice changes to water you have a positive heat of fusion. So...\n\n(12.37 g)(80 cal/g) = 989.6 cal\n\nNext you have to raise the water temperature from 0 degrees C to 100 degrees C. So...\n\n(12.37 g)(1.0 cal/deg/g)(100 deg) = 1237 cal\n\nNext you have the positive heat of vaporization.\n\n(12.37 g)(540 cal/g) = 6679.9 cal\n\nNext once it is steam, you must raise the temperature from 100 degrees C to 146 degrees C. So...\n\n(12.37g)(.48 cal/deg/g)(46 deg) = 273.1296 cal.\n\nNow that you have the energy for each step add them together.\n\n207.816 + 989.6 + 1237 + 6679.9 + 273.1296 = 9387.45 cal\n\nI'm pretty confident in my breakdown but it defintely wouldn't hurt to have someone else check it
945411
1 calorie is the amount of energy required to raise a gram of water 1k. 1k=1.8F.\n\nSo if you're increasing from 38 to 98.6F that's an increase of 60.6F= 33 and 2/3 K.\n\nSo total energy needed is 33.66 calories/gram.
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
882396
ummmm, maybe increase by a factor of 5 or so. Its an exothermic reaction to begin with and you're not limited by the amount of water( i think, if not not all your drano will dissolve) so it should be about five times the energy released compared to just two grams.
1117920
It will be water at 100 degrees that will stay that temperature longer b/c of the increased volume.  The only way to increase the temp is to increase the energy (temperature) while keeping the same mass (like putting flame under the container of water).  Since you are increasing mass AND energy, it balances out.
646478
Also watts, joules, and occasionally Calories (amount of heat needed to raise one liter of water one degree centigrade.)