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How do scientists determine how many calories a food has?
They breakdown the food to its simplest form then place it in a metal cup which is in a water box. After they place the cup in the water they take the temperature of the water (Tinitial). When they have the temperature of the water they ignite the food and check the temperature after the fire(Tfinal). Then they use the equationq=Masswater x specific heatwaterx ΔTwater(q=Heat in calories or joules) (T=temperature) (ΔTwater=Tfinal-Tinitial) (spefic heat of water= 4.18 J/g °C) (Masswater= Mass of water in box - mass of box w/o water)(joules to calore conversion 1 calorie=4.184 Joules)
What unit of measurement is used to record heat transfer?
Heat transfer is measured in Joules. At times, teachers look for calories, so the conversion for Calories to Joules is 1 calorie = 4.184 joules. The formula used for heat transfer is q = m(T)C .... q is amount of heat transferred, m is mass, T is the change in temperature, C is the specific heat.
Specific heat capacity of water?
The specific heat of water is 4.184 J/g*°C, which is read 4.184 Joules per gram degree Celsius. It can also be stated as 1.00 cal/g*°C, which is read calories per gram degree Celsius (same as centigrade). These values mean that it takes 4.184 Joules of energy to raise the temperature of 1 gram of water 1 degree Celsius. Or it takes 1.00 calories of energy to raise the temperature of 1 gram of water 1 degree Celsius. 4.184 Joules = 1.00 calorie
How do units that are used to measure heat are different from the units that are used to measure temperature?
Heat is measured in calories or BTU, whilst temperature is measured in degrees Celsius or Fahrenheit.1 calorie is the heat to raise 1 gram of water by 1 degree C1 BTU is the heat to raise 1 pound of water by 1 degree FHeat is the amount of energy in a system measured in Joules. Temp is the MEASURE of the AVERAGE molecular motions in a system F/CAnother AnswerIn SI, heat is measured in joules and temperature in kelvin.
Approximately how many Joules of heat are needed to completely change 10.0 grams of ice to water at the melting point temperature?
To completely change 10.0 grams of ice to water at the melting point temperature, we need to calculate the heat required for the phase change from solid to liquid and the heat needed to raise the temperature of the resulting water to the melting point temperature. The heat of fusion for water is 334 J/g, so the heat needed for the phase change is 10.0 g * 334 J/g = 3340 J. The heat needed to raise the temperature of the resulting water to the melting point temperature is calculated using the specific heat capacity of water, which is 4.18 J/g°C. The temperature change is from 0°C to 0°C, so no additional heat is needed for this step. Therefore, the total heat required is 3340 J.
How many watts to heat 1 liter of water?
It takes about 4.18 Joules of energy to heat 1 gram of water by 1 degree Celsius. Therefore, to heat 1 liter (1000 grams) of water by 1 degree Celsius, it would require about 4180 Joules. Converting this to watts depends on the time taken to heat the water.
How much energy does it take to heat water from 33degrees F to 34degrees F?
The specific heat capacity of water is about 4180 joules per (kg * kelvin), so you need the mass of the water, and the temperature rise in kelvin (or degrees celcius) example: 1 kg water raised 33 to 34 deg. F ( 0.555 deg. celcius) so, 1 * 0.555 = 0.555, then 0.555 * 4180 = 2320 joules
What is is the amount of heat needed to raise the temperature of one kilogram of water one degree centigrade.?
The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius. To raise the temperature of one kilogram (1000 grams) of water by one degree Celsius, it would require approximately 4180 Joules of heat energy.
How many calories required to raise temperature of 1 liter of water 1 degree centigrade?
Well, let's see. Water at room temperature has a heat capacity of 4.18 J/g-C, and water also has a density of 1g/mL. If there's one litre of water, there's 1000 g of water. If the change in temperature is 1 C, and there are 1000 g of water, and specific heat capacity's 4.18... Q = mcT Q = (1000g)(4.18J/g-C)(1 C) Q = 4180 J So you need 4180 J of heat. *************************************** The definition of one calorie is as stated in the question. One calorie is equivalent to 4.18 Joules. Not sure where the maths went wrong but just so you know.
What is the heat of vaporization of water in joules per kilogram?
The heat of vaporization of water is 2260 joules per kilogram.
What is the latent heat of vaporization of water in joules per kilogram?
The latent heat of vaporization of water is 2260 joules per kilogram.
Suppose you want to heat 40g of water by 20c how many joules of heat are required?
334.8 Joules
How much energy is required to raise the temperature of one liter of water 1 degree centigrade?
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.
How many joules are required to boil one liter of water?
Any number of joules, no matter how small, will raise the temperatureof the water. The total number required in order to accomplish the jobdepends on the final temperature you want to see. The higher that is,the more energy it will take to reach it.
What is the specific heat of water in joules per kilogram degree Celsius?
The specific heat of water is 4186 joules per kilogram degree Celsius.
How long does it take to heat one liter of water from a heat source?
It depends on the how many degrees you wish to change the water and the wattage of the heater. Obviously a 1500 watt heater will do it faster than a 1000 watt heater. You might want to begin by looking at the heat transfer formula: heat in joules equals mass times change in temperature times specific heat of the material (water in this case).
How much energy would be required to raise one liter of water by one degree celsius?
280,000 gallons * 16 pounds per gallon = 4,480,000 pounds of water 1 BTU to raise 1 pound of water 1 degree F (at 60 to 61, Standard Atmospheric Pressure) 1 pound of propane ~ 22,000 BTU 4,480,000 / 22,000 ~ 203 203 pounds of propane ~ 50 gallons Burning approx 50 gallons of propane will heat 280,000 gallons of water by one degree. (Hardly a geothermal event, they burn more propane in the average tailgate at a NFL game.)
