Use this formula.
q(in Joules) = Mass * specific heat * change in temperature
I will use specific heat of water at 25 C. You can look up specific heat of steam. You say " heat to " so I assume you have final and initial heat backwards.
q = 25.0 grams H2O * 4.180 J/gC * (100.0 C - 29.25 C )
q = 7393 Joules
The specific heat capacity of water is 4.18 Joules/gram degrees Celsius. Therefore, it would take 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
To melt 1 gram of ice at 0°C, it requires 334 joules of energy. So for g grams of ice, the energy needed would be g multiplied by 334 joules.
The specific heat capacity of water is 4.186 J/g°C. Since there are 1000 grams in a kilogram, it would require 20,930 Joules of energy to increase the temperature of a kilogram of water by 5 degrees Celsius.
It takes 4.184 joules of energy to change the temperature of 1 gram of water by 1 degree Celsius.
The amount of energy needed to increase one gram of water by one degree Celsius is known as the specific heat capacity of water, which is 4.18 Joules/gram °C.
Approx 2940 Joules.
The specific heat capacity of water is 4.18 Joules/gram degrees Celsius. Therefore, it would take 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
8.200 J
To melt 1 gram of ice at 0°C, it requires 334 joules of energy. So for g grams of ice, the energy needed would be g multiplied by 334 joules.
I will use this formula. Some conversion will be required. ( I only know specific heat iron in J/gC ) q(Joules) = mass * specific heat * change in temperature Celsius 3 kilograms cast iron = 3000 grams q = (3000 g)(0.46 J/gC)(120 C - 30 C) = 124200 Joules (1 kilojoule/1000 joules) = 124.2 kilojoules of energy needed ===========================
539 calories per gram for heat of vaporization plus 1 cal/gram/degree C 100 degrees C - 80 degrees C = 20 degrees C (539 calories + 20 calories) X 50 kg X 1000 gm/kg = 27950000 cal = 27,950 kcal
The specific heat capacity of water is 4.186 J/g°C. Since there are 1000 grams in a kilogram, it would require 20,930 Joules of energy to increase the temperature of a kilogram of water by 5 degrees Celsius.
The amount of energy needed to raise the temperature of a substance is calculated using the formula ( Q = mc\Delta T ), where ( Q ) is the heat energy (in joules), ( m ) is the mass of the substance (in kilograms), ( c ) is the specific heat capacity (in joules per kilogram per degree Celsius), and ( \Delta T ) is the change in temperature (in degrees Celsius). This formula helps determine how much energy is required to achieve a desired temperature increase for a given mass of a substance.
It takes 4.184 joules of energy to change the temperature of 1 gram of water by 1 degree Celsius.
The specific heat capacity of water is 4.18 J/g°C, which means it takes 4.18 Joules of energy to raise 1 gram of water by 1 degree Celsius. To raise the temperature of 8.1 g of water by 20 degrees Celsius, you would need 8.1 g * 20°C * 4.18 J/g°C = 676.56 Joules of energy.
The amount of energy needed to increase one gram of water by one degree Celsius is known as the specific heat capacity of water, which is 4.18 Joules/gram °C.
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.