226,ooo j
When 2 grams of water is converted from liquid to gas at its boiling point, it requires an input of approximately 2260 joules of energy. Conversely, when 2 grams of steam condenses into liquid water, it releases around 2260 joules of energy.
The change in temperature is 21 degrees Celsius. To calculate the energy required, we use the formula: Energy = mass * specific heat * change in temperature. Plugging in the values, Energy = 1.3g * 0.131 J/g°C * 21°C = 35.247 Joules. Therefore, 35.247 Joules of energy is required to heat 1.3 grams of gold from 25°C to 46°C.
The energy needed to change ice into water is called the heat of fusion. For ice, this value is around 334 joules per gram. So, for 3 grams of ice, the energy gained when it changes to water would be around 1002 joules (334 joules/gram * 3 grams).
To solve this you would use the equation q=mCs deltaT, in which q is the amount of energy needed for the reaction (in Joules), m is the mass of the substance in grams, Cs is the specific heat, and deltaT is the change in temperature of the reaction. In this case, q = what you are looking for Cs = 4.184J/g*C m = 46.0 grams delta T = 100.0 degrees celsius q = 46.0g x 4.184j/g*c x 100.0 *c q = 19246.4 Joules
To calculate the energy required to heat water, you would need to know the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C. Assuming we are heating the water by 1°C, the energy required would be 46.0g * 4.18J/g°C * 1°C = 192.28 Joules.
226,ooo j
When 2 grams of water is converted from liquid to gas at its boiling point, it requires an input of approximately 2260 joules of energy. Conversely, when 2 grams of steam condenses into liquid water, it releases around 2260 joules of energy.
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
The change in temperature is 21 degrees Celsius. To calculate the energy required, we use the formula: Energy = mass * specific heat * change in temperature. Plugging in the values, Energy = 1.3g * 0.131 J/g°C * 21°C = 35.247 Joules. Therefore, 35.247 Joules of energy is required to heat 1.3 grams of gold from 25°C to 46°C.
0000000000000000.11 joules
Grams Liquid x mol/g x Hfusion
5 grams of protien
To get ice to its melting point, it must absorb 334 joules of energy per gram of ice. So, for 150 grams of ice, the total energy required would be 50,100 joules (334 J/g * 150 g).
The energy needed to change ice into water is called the heat of fusion. For ice, this value is around 334 joules per gram. So, for 3 grams of ice, the energy gained when it changes to water would be around 1002 joules (334 joules/gram * 3 grams).
Voltage is electrical pressure and Joules is a power rating so other info is needed
A liquid (or fluid) ounce is a measure of volume. A gram is a measure of mass. Therefore, the number of grams per fluid ounce of a liquid depends on what the liquid is, specifically on what its density is.
In principle, those two things measure different things (mass, and energy). To get the mass associated with a certain amount of energy, divide the number of joules (1 kJ = 1000 joules) by the square of the speed of light, i.e., divide by 9 x 1016 (m2/sec2). The answer is in kilograms; you an multiply this answer by 1000 to convert into grams.