2,641,760J...
If its in Celsius then another 13 degrees are needed because water boils at 100 degrees Celsius
to solve this we use the formula Q(heat) = mc(change in temp) so, Q=(10g)(4.19J/gC)(18-22) Q=-167.6kJ of heat.
The boiling point of water is 100 degrees Celsius and the melting point of water is 0 degrees Celsius
It is warm because water freezes at 0 degrees Celsius and water boils at 100 degrees Celsius
114.14g NaCl
2,641,760J...
The number of calories required will depend on the mass of water which is to be heated.
42 J
there are no calories in water you idiot
(5)(3)= 15 calories. 1 calorie is the energy (heat) to raise 1 gram of water by 1 degree celsius, so 5 grams of water (3 degrees Celsius) = 15.
it can be made to boil at 105 degrees Celsius if we add impurities to it,as impurities raise the boiling point.
Q=mcΔT Q=14 x 4200 x 21.6 Q=1270080J
If its in Celsius then another 13 degrees are needed because water boils at 100 degrees Celsius
4,200 J/kgC (83-4) x 8kg x 4,200 = 2,654,400 joules
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
12 degrees Celsius
12 degrees Celsius