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Joules needed to rise timperature of 2.0l of warer from 20degrees celsius to 100 degrees celsius?
To calculate the joules needed to raise the temperature of water, you can use the formula Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, calculate the mass of 2.0 liters of water in grams (1 mL of water = 1g), then use the specific heat capacity of water (4.18 J/g°C) to find the total heat energy required. The final step is to calculate the joules needed using the formula, remembering that 1 calorie is equivalent to 4.18 joules.
What is the amount of heat needed to raise that temperature of 1 kilogram of water 1 degree Celsius?
The amount of heat needed to raise the temperature of 1 kilogram of water by 1 degree Celsius is 4186 Joules, which is the specific heat capacity of water.
The temperature of 15 grams of water is increased by 3.0 Celsius degrees How much heat in Joules was absorbed by the water?
To calculate the heat absorbed by the water, you can use the formula: heat = mass * specific heat capacity * temperature change. First, determine the specific heat capacity of water (4.18 J/g°C). Then, plug in the values: heat = 15 g * 4.18 J/g°C * 3.0°C. The heat absorbed by the water is 188.1 Joules.
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 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
Joules needed to rise timperature of 2.0l of warer from 20degrees celsius to 100 degrees celsius?
To calculate the joules needed to raise the temperature of water, you can use the formula Q = mcΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, calculate the mass of 2.0 liters of water in grams (1 mL of water = 1g), then use the specific heat capacity of water (4.18 J/g°C) to find the total heat energy required. The final step is to calculate the joules needed using the formula, remembering that 1 calorie is equivalent to 4.18 joules.
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.
What do you call the amount of energy needed to change a material from a solid to a liquid. propane or water?
The amount of energy needed to change a material from a solid to a liquid is called the latent heat of fusion. This value differs for different substances, but for water, it is 334 joules per gram, while for propane it is 93 joules per gram.
How much energy is needed to melt 0.25 moles of water?
The necessary heat is 9,22 joules.
How many joules of heat are needed to completely vaporize 24.40 grams of water at its boiling point?
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
How many joules are required to boil 21.1 g of water at 100 C?
The heat required to boil water can be calculated by multiplying the mass of water (21.1 g) by the specific heat capacity of water (4.18 J/g°C) and the temperature change (100°C - initial temperature). This calculation results in 8.82 kJ or 8820 J of energy needed to boil 21.1 g of water at 100°C.
How many joules of heat are needed to raise the temp of 2.83 kg of water from 38C to 46C?
0.0796
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 heat change when 55.0g water cools from 60 degree celsius to 25.5?
To calculate the heat change, you need to use the formula: q = mcΔT. First, calculate the specific heat capacity of water (4.18 J/g°C). Then calculate the change in temperature (60°C - 25.5°C). Finally, substitute these values into the formula along with the mass of water (55.0g) to find the heat change in joules.
What is the amount of heat needed to raise that temperature of 1 kilogram of water 1 degree Celsius?
The amount of heat needed to raise the temperature of 1 kilogram of water by 1 degree Celsius is 4186 Joules, which is the specific heat capacity of water.
How much heat in joules is needed to raise the temperature of 4.0 L of water from 0 degrees Celsius to 70.0 degrees Celsius?
It takes 4.186 Joules to heat one gram of water by 1-degree Celsius. 4.186 * 4000 = 16,744 Joules to heat 4 kilos of water by 1-degree. 16,744 * 70 = 1,172,080 Joules. The above assumes that one litre of water weighs exactly 1 Kilogram.
The temperature of 15 grams of water is increased by 3.0 Celsius degrees How much heat in Joules was absorbed by the water?
To calculate the heat absorbed by the water, you can use the formula: heat = mass * specific heat capacity * temperature change. First, determine the specific heat capacity of water (4.18 J/g°C). Then, plug in the values: heat = 15 g * 4.18 J/g°C * 3.0°C. The heat absorbed by the water is 188.1 Joules.
