Q = MC(delta T). So, the change in temperature is equal to Q/MC. This expands to 259/(25 * 4.18). The final temperature is 24.48 degrees Celsius.
The increase in temperature can be measured by the following formula Q=mass*specific heat capacity*(change in temperature). 5000=100*4.18*(OT-20). By solving for OT, we get the answer as 32 degree celcius.
Using the formula for transfering heat: q=mc(dT), where
q = heat
m = mass of the water
c = specific heat capacity of water (4.184 J/g.oC, or 1 cal/g.oC)
dT = "delta" T, the change in temperature, T2-T1
q=(100g)(4.184J/g.oC)(82oC - 25oC) = (418.4J/oC)(57oC) = 24,000 J, or 24kJ
[Note: There are only 2 significant figures in 25oC and 82oC, so the answer can only have 2 significant figures.]
q=mcdeltat
25,100J=100g*4.18*deltaT
25+deltat=answer
At a pressure of 1 atmosphere, it is approximately 1500 small calories or gram calories (cal). This is equivalent to approx 1.5 large calories or food calories (Cal).
53.8 degrees
150 gm of water at 60 degrees
24.5
700
8.62
1200 cal
No. The quantity of energy required to raise the temperature of water is different depending on the phase of water. This is especially true at or near a phase transition as thermal energy is absorbed during a phase transistion thus altering the amount of energy required to raise the temperature of said water.
1 kilocalorie
700
The answer is 2 calories.
1 calorie is needed to raise 1 g of water 1 °C. 350 * 22 = 7700 calories ■
A Calorie is defined as the amount energy required to raise 1 gram of water by 1 degree Celsius. 79.7cal are required to a phase change in 1 gram of ice to water. It is being assumed that the temperature of the ice is going to be 0 degrees Celsius, and not any colder. IE, the temperature of the water after its phase change to ice. So, since we have 60 grams of water, it will take (60*79.7) 4782cal just to complete the phase change of ice to water (cause the water to melt). If the ice measured 0 deg. Celsius, it still measures 0 deg. Celsius at this point. Now lets raise it by 1 deg. C. 60 Grams of water * 1 cal/degree C rise = 60 calories are required to raise 60 grams of water by 1 deg. C. 4782cal + 60cal = 4842 calories total to cause 60 grams of ice to phase change into water AND cause the water temperature to rise by 1 degree Celsius.
Depends on how high you want to raise the gram of water ;).
21 Kg = 2100 grams to rise the temperature of this amount of water by 2 degrees Celsius you need 2*2100 = 4200 calories or 17572.8 Joules.
23joules
1 calorie is the energy required to raise 1 gram of water by 1 degree C. So it would take 5 calories to raise it by 5 degrees C.
Energy required to raise 1 gramme of water by 1 degree C = 1 calorie also, 1 calorie = 4.186 Joules
The specific heat of water is 4.179 Joules per gram per degree Centigrade. The density of water is 1 gram per cubic centimeter, so one liter is 1000 grams. This means it takes 4179 Joules to raise one liter one degree Centigrade.
42 J
8.62