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Cooling or Heating First of all we have 3 phases, Gas, Liquid and Solid to change from gas to liquid to solid, cooling is required and to change from solid to liquid to gas, heating is required This type of heat is called latent heat, always think of water as an example Ice --> Water , heating Water--> Water Vapour, more heating
The heat released can be calculated using a calorimeter. Measure the initial and final temperatures and use the equation Q=mc(change in T). Where Q is heat, m is mass, and c is heat capacity.
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
In the case where heat is being added to generate steam from a container of water, as long as there is water still in the container, the temperature remains constant. It takes energy for water to change state into steam and all of the heat added goes to performing this task so the temperature stays the same, 212 degF (100 degC) at atmospheric pressure. While the water is boiling, the steam is "saturated," meaning any loss of heat would cause some of the steam to condense back into water. Once all of the water has been boiled and changed state, any additional heat supplied will cause the temperature to increase and the steam is "superheated," that is above its saturation temperature.
some of the energy,however, is absorbed by the land and water and change into heat
419.1 Joules are required to heat one gram of liquid water from 0.01 degC to 100 deg C. So the answer is 419.1*46 = 19278.6
Water has a specific heat of 4.18J/gC, so set up the specific heat equation: C (spec. heat)=q (joules)/mass x temp. change, so: 4.18 (spec. heat of water) = q/46g(100deg), so q = 4.18(4,600) = 19,228 joules (or 19.228 kJ).
The necessary heat is 0,155 kJ.
For an approximate calculation: specific heat capacity for water = 4.18 J/(g*degC) (how much energy is required per gram per change in degrees C) mass = 25g Change in temperature = 60-10 = 50 degC energy required = mass * change in temperature * specific heat capacity = 25g * 50 degC * 4.18 J/(g*degC)
The energy is 103,6 kcal.
How can the temperature DECREASE from 30.5ºC to 35.6ºC? That's an INCREASE in temperature. So, assuming you meant the temperature INCREASED to 35.6ºC, then it is an endothermic reaction.q = mC∆T = (1000 g)(4.184 J/g/deg)(5.1 deg) = 20,920 J = 20.9 kJ. This is the heat change for this reaction.
The metric unit for heat is the calorie - the heat required to raise 1 gram of water by 1 deg C. In the SI unit system it would be the kilocalorie - the heat to raise 1 kg by 1 degC
To change the temperature of water from 27ºC to 32ºC will depend on the mass of water that is present. Obviously, the more water, the more heat it will take. This can be calculated as follows:q = heat = mC∆T where m is the mass of water; C is sp. heat = 4.184 J/g/deg and ∆T is 5ºC (change in temp).
The needed heat is:Q = 10 x 20 x 0,031 = 6,2 calories
1 cal/gdegC x 225g x (100 - 50.5)degC heat to the boiling point 540 cal/g x 225g heat to vaporize the water 0.5 cal/gdegC x 225 g x (133-100)degC heat to chang temp of steam = 11137.5 cal + 121500 cal + 3712.5 cal = 14850 cal
one calorie of heat is able to raise one gram of water one degree Celsius so 400 calories could raise 1g of water 400 degrees, so it would raise the 80g by(400/80) 5 degrees Celsius plus the initial temp of 10 degrees, the 80g of water would have a final temp of 15 degrees Celsius
heat will change it