To convert 3 kg of ice at 0 degrees Celsius to water at the same temperature, you would need to supply 3,360,000 joules of heat (latent heat of fusion of ice is 334,000 J/kg). This energy will help melt the ice into water at 0 degrees Celsius.
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.
To convert Kelvin to Celsius, you subtract 273. So, 362 K is equal to 89 degrees Celsius.
Liquid water has a temperature range of 0 to 100 degrees Celsius. At 0 degrees Celsius, water freezes into ice, and at 100 degrees Celsius, water boils into steam.
Temperature is easy to convert from Fahrenheit to Celsius yourself. You can use the formula Tc = (5/9)*(Tf-32) where Tc = temperature in degrees Celsius, Tf = temperature in degrees Fahrenheit. 99 F is 37 C.
The boiling point of water is the temperature at which the vapor pressure of water is equal to the atmospheric pressure. As long as there is liquid water present, the temperature will remain at the boiling point (100 degrees Celsius at sea level), because the energy is being used to convert the liquid water into vapor rather than increasing the temperature.
100 degrees Celsius
heat energy required to raise the temperature of ice by 29 celsius =specific heat capacity of ice * temperature change *mass of ice + to change 1kg of ice at 0 celsius to water at 0 celsius =specific latent of fusion of ice*mass of water + heat energy required to raise the temperature of water by 106 celsius =specific heat capacity of water * temperature change *mass of ice + to change 1kg of water at 106 celsius to steam at 106 celsius =specific latent of fusion of ice*mass of steam
the freezing point of water is 0 degrees C, the corresponding temperature on the Kelvin scale is 273.16. Essentially one just subtracts 273.16 from the celsius reading to convert it into Kelvin.
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.
Can you help
The amount of energy required to raise the temperature of one gram of water by one Celsius degree is called the specific heat capacity of water. It is approximately 4.18 joules per gram per Celsius degree.
To convert Kelvin to Celsius, you subtract 273. So, 362 K is equal to 89 degrees Celsius.
The specific heat capacity of water is 4186 J/kg*C. To calculate the heat required, use the formula: heat = mass * specific heat capacity * change in temperature. Plugging in the values, the heat required to raise the temperature of 0.25 kg of water by 10 degrees Celsius is approximately 1046.5 Joules.
0 degrees Celsius is the temperature at which water will freeze. This system of measuring temperature uses water as a basis where 0o Celsius is the temperature at which water freezes and 100o Celsius is the temperature at which water boils.
The simple beginning is that the definition of a calorie is "the energy required to raise the temperature of 1g of water 1°C." Therefore, the energy required to raise 17g of water 32°C: 17*32=544 cal. However, the question asked about ice. There is an extra bit of energy required for the change of physical state. The energy required to convert 1 gram of ice at 0°C to liquid water at 0°C is called the "latent heat" and is equal to about 80 cal. To convert 17g of ice, we multiply this together: 17g * 80cal/g = 1360 cal. So, we add this energy required for the change of state to the energy required to raise the listed quantity to the required temperature and we get 544 cal + 1360 cal = 1904 cal, assuming no heat is lost to the environment. I hope this clarifies some things.
The temperature in Celsius at which water boils is 100 degrees.
The answer is 323.15 K (approx.). The Kelvin scale starts at absolute zero and is used in scientific laboratories. Celsius is for general use and set 0 and 100 as melting and boiling point of water respectively. To convert from C to K, add 273.15 to C.