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How many SI calories does it take to raise one liter if water 6 degrees Celsius?
It takes 6 SI calories to raise one liter of water by 6 degrees Celsius.
What liquid is -100 Celsius?
Liquid nitrogen is a common substance that can exist at temperatures around -196 degrees Celsius, well below -100 degrees Celsius. At -100 degrees Celsius, substances like liquid oxygen or liquid argon may be present, depending on the specific conditions and composition.
What is 4 trillion degrees Celsius in Fahrenheit?
4 trillion degrees Celsius = 7.2 × 1012 degrees Fahrenheit.
How many calories are required to heat 37.4g of lead from 15 to 60 degrees Celsius?
This heat is 51, 33 cal.
What temperature in Celsius is 13 degrees warmer than -5 degrees Celsius?
18
How many calories are released during the process of condensation?
To transform 1 gram of ice at 0 degrees Celsius to 1 gram of water vapor at 100 degrees Celsius, 720 calories are added (absorbed). There are no calories released during the process.
How many calories are required to heat 38.2 grams of aluminum from 102 degrees celsius to 275 degrees celsius?
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
How many SI calories does it take to raise one liter if water 6 degrees Celsius?
It takes 6 SI calories to raise one liter of water by 6 degrees Celsius.
Heat is measured in Celsius degrees true or false?
False. Heat is measured in units of energy such as calories or joules, not in degrees Celsius. Temperature, on the other hand, is measured in degrees Celsius.
How many calories are needed to convert 12.5 grams of ice at 0 degrees Celsius to steam at 100 degrees Celsius?
To convert 12.5 grams of ice at 0 degrees Celsius to steam at 100 degrees Celsius, you would need to provide heat energy for three main processes: heating the ice from 0 degrees Celsius to 100 degrees Celsius, melting the ice into water at 0 degrees Celsius, and then heating the water from 0 degrees Celsius to steam at 100 degrees Celsius. The total calorie requirement would be determined by the specific heat capacities and heat of fusion and vaporization of water.
4 kg of ice at 0 degrees Celsius is converted to steam at 100 degrees Celsius How much heat is needed?
To convert 4 kg of ice at 0 degrees Celsius to steam at 100 degrees Celsius, you would need to calculate the heat required to raise the temperature of ice to 0 degrees Celsius, melt the ice to water at 0 degrees Celsius, raise the temperature of water to 100 degrees Celsius, and then convert water to steam at 100 degrees Celsius. The total amount of heat needed can be calculated using the specific heat capacities and latent heats of fusion and vaporization of water.
How many calories to raise i gram of water 40 degrees c?
To raise 1 gram of water by 40 degrees Celsius, you need 40 calories. This is based on the specific heat capacity of water, which is approximately 1 calorie per gram per degree Celsius. Therefore, for a temperature increase of 40 degrees, you multiply 40 degrees by 1 calorie per gram.
Is temperature measured in calories?
No, temperature is not measured in calories. Temperature is typically measured in degrees Celsius (°C) or degrees Fahrenheit (°F), while calories are units of energy related to the amount of heat required to raise the temperature of one gram of water by one degree Celsius.
What is the specific heat of 40g of a substance if it requires 195.3 calories of energy to raise its temperature from 5 to 15 degrees Celsius?
488.25 J/kg/*C
If 1 gram of iron requires 0.11 calories to raise the temperature 1 degree Celsius what is iron's specific heat?
0.11cal/g degrees C
How many calories are needed to change 1 gram of 100 degrees Celsius boiling water to 100 degrees Celsius steam?
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
If silver has a specific heat of about 0.06 calg degrees Celsius how many calories will it take to raise the temperature of a silver spoon that weighs 40 g from 20 to 88 degrees Celsius?
use the equation q=mc∆t, where q is the calories required, m is mass, c is specific heat capacity and ∆t is the change in temperature. Therefore... q=(40g)(0.06 cal/g◦c)(88c - 20c) q=163.2 calories
