Q=m s dT m = s dT/Q = 0.00924 X (100-25.3)/125 = 0.0055122 g
1.18 x 10^4 J
The heat is 11,77 kJ.
6 Calories
there are no calories in water you idiot
quantity of heat required =mass*specific heat of water*change in temperature Q=160*1*(77-19)=9280 calories=9.28 kilo calories
250g of water at 10C needs to lose 1 cal/g/C or 2500 calories to drop temperature to zero. The latent heat of fusion of water is 80 calories per gram at 0C so, the water needs to lose 20,000 calories to turn to ice at 0C Finally, the ice needs to lose 0.316 cal/g/C or 790 calories to drop to -10C The total heat released is then 2500 + 20,000 + 790 = 23,290 calories
1.7293
To transform 1 gram of ice at 0 degrees Celsius to 1 gram of water vapor at 100 degrees Celsius, 720 calories are added (absorbed). There are no calories released during the process.
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
The number of calories required will depend on the mass of water which is to be heated.
488.25 J/kg/*C
0.11cal/g degrees C
use the equation q=mc∆t, where q is the calories required, m is mass, c is specific heat capacity and ∆t is the change in temperature. Therefore... q=(40g)(0.06 cal/g◦c)(88c - 20c) q=163.2 calories
140 degrees Celsius is 140 degrees Celsius regardless of the specific location of measurement.
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
80.5 calories 35-12=23 23*3.5=80.5 1c raises 1 gr. h2o 1degree centigrade Here is the formula, it should help a lot:Total Number of Calories = (Specific Heat of Water) ×(Mass of Water) × (Absolute Temperature Change)
115.2
There are 1000 g in one liter Use the formula Q=mc(delta)T Q=1200g(1calorie/g degrees celsius)(70 degrees - 20 degrees) Q=60,000 calories Q=60 kilocalories
6 Calories