Heating of water=m x s x delta T,where m is the mass ,s is the specific heat of water(1 cal/gm)=5x1x(50-25)
=125 cal
The energy required to raise the temperature 1 degree Celsius of 1 gram of water (1 mL) is 1 calorie (=4.18 J). So for 1 kg, 1Kcal (= 4180 J = 4.18 KJ) is required. To raise it 60 degrees, just multiply by 60 and for 10 kg multiply by 10 again. That would make 2.508 MJ (= 2508000 J) Now this is not completely accurate. The energy required to raise the temperature of water differs at 20 degrees from that at 60 degrees. The difference is small (~0.05 J or something like that) but still present.
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.
A change in water temperature with no change of state requires one calorie per gram of pure water per degree Celsius. 90° - 15° = 75° 80 g x 75° C x 1 cal/g/°C = 6000 cal = 6 kcal
That's going to depend on how much water you're responsible for. Teacup at 60 degrees . . . very few BTU. Swimming pool at 60 degrees . . . many more BTU. It's also going to depend on whether you're talking about Celsius or Fahrenheit degrees. Fahrenheit degrees . . . fewer BTU. Celsius degrees . . . more BTU. (Also, the water will escape as you pass 100.) In general, one BTU is approximately the energy required to raise the temperature of 1 pound of water 1 degree Fahrenheit. You can take it from there, when you reach the job site and determine the exact scope of the work.
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 number of calories required will depend on the mass of water which is to be heated.
42 J
15.37684 joules
2,641,760J...
2,641,760J...
Q=mcΔT Q=14 x 4200 x 21.6 Q=1270080J
3.50 J
raise the temperature of the body by 1 Celsius
1935 J (apex)
10-12
15480.80