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.
Conversion needed.
4 kg (1000 grams/1 kg)
= 4000 grams
----------------------now,
q(thermal energy in Joules) = mass * specific heat * change in temperature
q = (4000 grams)(4.180 J/gC)(75 C - 25 C)
= 8.4 X 105 Joules
--------------------------
Approx 2940 Joules.
Approx. 600 - 800 degrees C / 1,100 - 1,500 degrees F
4,200 J/kgC (83-4) x 8kg x 4,200 = 2,654,400 joules
1 calorie increases 1 gram of water by 1 degree celsius. 4.18 Joules are needed to increase the temperature of 1 gram of water by 1 degree celsius. To reduce the 1 gram of water 1 degree celsius it would have to give off 1 calorie of energy. To calculate the energy multiply the mass in grams of water by 4.18 and by the change in temperature. The energy = 4.18 x m x change in T. The answer is in Joules. If you are using calorie as the unit of energy, replace 4.18 J by 1 C. Note that food is measured in kilocalories (Calories) not metric calories.
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
8.200 J
c(*) x mass x delta (change in temp)=joules4186x1.92(**)x80o =joules642969.6 joules
Approx 2940 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 ===========================
E = mass x specific heat x Δ°T Δ°T = new temperature - original temperature where Δ°T is equal to temperature change (Celsius in this case). The specific heat of Al is 0.900 J/g°C. Before we proceed to find the quantity of heat in joules, we must first find the temperature change. To calculate the temperature change, we must subtract the original temperature from the new temperature. Δ°T = 50°C - 25°C = 25°C In order to find the quantity of heat (joules), we must multiply mass, specific heat, and the temperature change (calculated above). E = 40.0g x 0.900 J/g°C x 25°C = 900 Joules or 9.0 x 102 Joules
mmmm enthalpy
The specific heat of air at 0 degrees Celsius is 1.01 Joules per gram or J/g. The specific heat of a substance is defined as the quantity of heat per unit mass needed to raise its temperature by one degree Celsius.
Approx. 600 - 800 degrees C / 1,100 - 1,500 degrees F
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
4,200 J/kgC (83-4) x 8kg x 4,200 = 2,654,400 joules
1 calorie increases 1 gram of water by 1 degree celsius. 4.18 Joules are needed to increase the temperature of 1 gram of water by 1 degree celsius. To reduce the 1 gram of water 1 degree celsius it would have to give off 1 calorie of energy. To calculate the energy multiply the mass in grams of water by 4.18 and by the change in temperature. The energy = 4.18 x m x change in T. The answer is in Joules. If you are using calorie as the unit of energy, replace 4.18 J by 1 C. Note that food is measured in kilocalories (Calories) not metric calories.
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