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The specific heat of water is 4.179 Joules per gram per degree Centigrade. The density of water is 1 gram per cubic centimeter, so one liter is 1000 grams. This means it takes 4179 Joules to raise one liter one degree Centigrade.
4186 Joules per liter per deg C. Not clear if we are raising the temperature BY 135 deg or TO 135 deg. So the answer is 4186 x 100 x rise in temperature. (Joules).
All objects have a specific heat capacity, the amount of energy it takes to increase the temperature of one gram of the object by one degree centigrade. Therefore, if an object has a specific heat capacity of say, 500 joules, and you wanted to heat 3 grams of that object by 1 degree, you would need 1500 joules of energy. But if you put that 1500 joules of energy into only 1 gram of the object, you would heat it by 3 degrees.
195 joule..
No, Fahrenheit is the Imperial Unit for temperature, not heat. Heat is energy in transit and is measured in joules (in the SI sytem).
Heat of vaporization of water is 2.26 x 106 joules per kg. Therefore 1 gram of water will need 2.26 x 103 joules.
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q = (20.5 g)(0.21 J/gC)(230o C- 30o C) = 861 Joules =========
The specific heat of water is 4.179 Joules per gram per degree Centigrade. The density of water is 1 gram per cubic centimeter, so one liter is 1000 grams. This means it takes 4179 Joules to raise one liter one degree Centigrade.
4186 Joules per liter per deg C. Not clear if we are raising the temperature BY 135 deg or TO 135 deg. So the answer is 4186 x 100 x rise in temperature. (Joules).
degrees and centigrade
334.8 Joules
q = mass * specific heat * change in temp. ( Joules, so conversion at end ) q = (93.6 g)(2.51 J/gC)[55o C - ( - 35o C)] = 21144.24 Joules ===============================
Heat is measured in unit of what...
70 calories per gram. (The specific heat capacity of water is 1 calorie per gram per degree C.) This could be converted into Joules if necessary using the conversion factor of 1 calorie = 4.18400 Joules.
Use the equation q=mc(delta t) (that is, heat equals mass times specific heat times the change in temperature) to answer the question. The specific heat of water is 4.186 Joules per gram-Celsius. Therefore, q=(40)(4.186)(20), which equals 3348.8 Joules of heat (or approximately 3.35 kiloJoules of heat).
1.25