3.50x10^4
To calculate the calories of heat available when the water cools to body temperature, you need to consider the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C. By using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change, you can find the answer.
water's specific heat is 1 calorie/mL/degree celsius. if you assume the density is 1 g/mL then we have 50 mL of material going from 100 to 37 degrees C. So take Mass X Change in Temp X Specific heat and that is your answer.
To vaporize 1 gram of boiling water at 100 degrees Celsius, it would require approximately 2260 joules, which is equivalent to about 0.54 calories. This energy is needed to break the intermolecular bonds holding water molecules together, allowing them to escape into the gas phase.
As the water vapor cools at 101 degrees Celsius, it will undergo a phase change and condense into liquid water. This is because the cooling causes the water vapor to lose energy and come together to form liquid droplets.
False. Water expands when it cools from 38 to 32 degrees Fahrenheit. However, it contracts as it freezes below 32 degrees Fahrenheit.
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.
250g of water at 10C needs to lose 1 cal/g/C or 2500 calories to drop temperature to zero. The latent heat of fusion of water is 80 calories per gram at 0C so, the water needs to lose 20,000 calories to turn to ice at 0C Finally, the ice needs to lose 0.316 cal/g/C or 790 calories to drop to -10C The total heat released is then 2500 + 20,000 + 790 = 23,290 calories
No, water's density decreases as it cools. Water reaches its maximum density at around 4 degrees Celsius, and as it cools further, the water molecules form a crystalline structure, causing the density to decrease.
To calculate the calories of heat available when the water cools to body temperature, you need to consider the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g°C. By using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the temperature change, you can find the answer.
1370 calories
It takes 6 SI calories to raise one liter of water by 6 degrees Celsius.
To calculate the energy released when the steam cools to water, you need to consider the specific heat capacity of water and steam. The equation Q = mcΔT can be used, where Q is the energy released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Once you have the energy released, you can convert it to joules.
Ittakes 1 calorie for every degree C change of 1 gram of water answer: 100
100
super-cooled
water's specific heat is 1 calorie/mL/degree celsius. if you assume the density is 1 g/mL then we have 50 mL of material going from 100 to 37 degrees C. So take Mass X Change in Temp X Specific heat and that is your answer.
A calorie is the amount of heat you need to raise the temperature of one gram of water by one degree Celsius. Assuming you are raising the temperature of the water from twenty degrees Celsius to ninety-nine degrees Celsius, it would take 20,000 calories. To calculate this, subtract 20 from 99. This is the amount of degrees you need to raise the temperature of the water by. Then multiply that number by 256, the amount of water in grams. You should get 20,244 calories. In significant digits, your answer should be 20,000 calories.