What is q (heat) when 12.0g of water is heated from 20(degrees Celsius) to 100(degrees Celsius).?
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Given:\nm <mass of substance> = 12.0g or 12/1000 kg\nc <specific heat capacity of water> = 4180J per kg per degree celsius\nt <change in temperature> = 20 degrees to 100 degrees i.e. 80 degree celsius\n\nFind:\nHeat req.\n\nSolution:\nHeat required or Q\nQ = m * c * t\n  = 12/1000 * 4180 * 80\n  = 12/10 * 418 * 8 (cancelling 0s of 4180&80)\n  = 12/5 * 209 * 8 (div. 418&10 by 2)\n  = 12/5 * 1672 (mult. 207&8)\n  = 20064/5 (mult. 12&1672)\n  = 4012.8\n\nAnswer = 4012.8 Joules or 4.0128 KJ
482741
You need a few different constants to figure this out.\n\nWater has a Specific Heat of 1 kcal/kg/°C\nThat means it needs 1 kcal per kilogram to heat the water 1 degree celsius\n\nIt has a density of 1000 kg/m3\n\nIt has a Latent heat of Fusion of 80 kcal/kg (melting) and 65 J/kg (freezing)\n\nSo if you know the volume you can work out the weight of the water. Once you have the weight (or you already know it), you can work out the rate of the drop (or increase) of water temperature.\n\nOnce you get to the point of freezing, or melting, simply add on the latent heat, and you can then get the total amount of energy required to change the water from one temperature to the next.\n\nSo for example;\n\n1) How much energy to heat 2 kg water from 18°C to 100°C is =>\n\n2kg * 1 * (100-18) = 164 kcal (The heat up curve is fairly straight)\n------------------------------->\n\n2) 2kg of water from 5°C to -10°C is as follows;\n\n(2 x 1 * (5 - -10)) + (65 * 2) = 160 kcal\n\nCurve drops down fairly constantly, then flats out until the water turns to ice, then continues to drop until you are at your temperature.\n\nI hope that helps a little.\n\nIf you want to figure out the times for all this to happen, you need to either know what heat source you are using (for example the wattage of the kettle element), and also you need to figure out the heat losses from the container (e.g. kettle or pipe, etc.)\n\nThese can all then be converted to an equivalent kcal number.
1008978
well this is really a chemistry problem but the formula is q=mC(delta) T, meaning heat= mass times the specific heat capacity times the change in temperature, so to set this up we use, 1.68^10 j = m (4.18 j/g) (80 degrees) becuase 4.18 is C and 80 degrees is the amount u wish to raise it, so then u get 1.68^10 j = 334.4m, divide both sides by 334.4 and you get .53558=m so your mass is .536g of water
800141
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
64882
Water's boiling point is 100 degrees Celcious (212 Fahrenheit), at zero altitude. Altitude can change the boiling point for water.\n\nYou can calculate that if you know your altitude, by following the link provided below.
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.
262451
Q=mst  \nwhere t is the change in temp.\n\n=5*200*s\n=1000*s\n\nwhere the value of 's' has to be specified
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1) Calculate the heat needed to melt 100 grams of ice completely: \nMass of ice * Heat of Fusion\n100g * 333J/g = 33300 J\n\n2) Calculate the cooling we can obtain \nMass of water * specific heat of water * temperature loss\n180g * 4.2 J/g K * dT = 33300 J => dT = 44.0 degrees\n\nConclusion we need only to melt 100g * 30/44 = 68.1 gram of ice to cool down the water to zero degree. 31.9 grams of ice remains.\n\n3) 1) Calculate the heat needed to melt 50 grams of ice completely: \nMass of ice * Heat of Fusion\n50g * 333J/g = 16650 J\n\n4) Calculate the cooling we can obtain \nMass of water * specific heat of water * temperature loss\n180g * 4.2 J/g K * dT = 16650 J => dT = 22.0 degrees\n\n5) Calculate the average temperature of the water.\nAfter melting of the ice we have 50 g of water at 0C and 180 g of water at 30-22=8C. This average out to 6.3C
554942
Q =  m C (T-t)\nQ = (2kg)(2100j/kg)(0°c-(-20°C))\n= 84000 j\nthen\nQ = mL\n= (2)(335,000)\n= 670000 j\n\nthen\nQ = m c (T-t)\n= (2)(4200)(100-0)°C\n= 840000\n\nthen\n\nQ = mL\n= (2)(2,260,000)\n= 4520000\n\nthen\nTotAl = 8400+670000+840000+4520000\n\n\n= 6,038,400 J
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Boiling water has a temperature of 100 Celsius, steam can have higher temperature.
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Get an electrical heating element, a DC power supply, a thermometer, a voltmeter and an ammeter.\nPass electricity through the heating element, dipped into a measured volume of water (or salt water) in an insulated cup, and measure the current, the voltage and the time taken. This will give you the heat energy supplied. Measure the temperature rise, and work out how much energy would be required to raise 1g of water (or salt water) by 1 degree Celcius. The answer should be close to 4.2!
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1. Water boils- 212 degree F/100 degree C\n2. Normal Body temp- 98.6 degree F/37 degree C\n3. Water Freezing temp- 32 degree F/ 0 degree C
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100 degrees celsius at sea level.
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Q=m.c.(delta)T\nQ=0.4 x 4200 x (20-16) = 6720 J\n\n6720=0.2 x c x (100-20) \nc= 420 J / kg K
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Energy required to heat 1liter of water:\n\nQ = V(liters) * d(kg/liter) * Cp(kJ/kg/K) * (Tfinal - Tinitial)\nwhere Cp is the specific heat capacity of water, which is temperature dependant but for which you can use an average value for the temperature range.\n\nQ = 1 * 1 * 4.18 * 73\nQ = 305.14 kJ\n\n1 kW.h = 3600 kJ\n\nQ = 305.14 kJ / 3600 (kJ/kW.h)\nQ = 0.085 kW.h\n\nPrice = 0.085 * 10 = 0.85 cents
836375
e. 650cal.\n\nq=mc(change in temp/ cal)\n  = 100 x 1 x 6.5 (since 31.5 - 25.0 = 6.5 deg C)\n  = 650cal.\n\nTake note of the SI unit for 'c'; it determines the necessity to convert temperature from deg C to K, or mass from g to kg.
887338
Check the thermometer - I think you will find it is 102 F - at 102C you would cook! The boiling point of water is 212 F or 100C. Your body temperature is normally 98.4 F and so you are in a room about 3.5 F above your normal body temperature - still warn though. the body prefers a lower temperature - so it can control body temperature through the means of perspiration.