Even More things to know about thermal energy:
~When temperature increases that means that the molecules have more KE.
~The total energy (TE) within an object is its internal energy
~Everything expands when heated and contracts when cooled. Then exception for this is water in the temperature range near freezing; ice expands from its liquid state.
~Heat is the the net energy transferred from one object to another because of temperature. So when you add heat the internal energy increases.
~Heat is measured in calories (cal) and kilocalories (kcal).
~1 cal is the amount of needed to raise 1g of pure water by 1˚C
~1 kcal is the amount of needed to raise 1kg of pure water by 1˚C
~a food calorie (Cal or C) is equal to 1kcal
~1C is equal to 100 cal (or 1kcal), which is equal to 4186J (approx. 4.2 kilojoules or kJ)
~Cal is the amount of energy produced when a given amount of food is burned
~Those units are all for Standard Units (SI). In British units, 1kcal= 4 British thermal units (or Btu)
~When heat is added, the temperature of the substance increase.
~The specific heat of a substance is the amount of heat needed to raise the temperature of 1kg of the temperature of the substance by 1˚C
~Water has a specific heat of
~Specific heat is measured in
~The higher the specific heat of a substance, the higher the heat must be to raise the temperature. Also, the substance has a higher capacity for heat.
~amount of heat needed to change temperature (H) = mass x specific heat (c) x temp change OR H=mc∆T
∆T stands for the change in temperature (T)
*NOTE- this equation is used for substances that are not changing phase. Objects that are changing phase don not change T and therefore must use a different equation.
1. How much heat in kcal does it take to heat 95kg of bathwater from 15˚C to 40˚C?
Step 1- write what you know.
m=95kg, ∆T=25˚C, c=1kcal/kg˚C, trying to find H
*The answer will be in kcal. If c was in J/kg˚C, then the answer would be in J
The kg's and ˚C's then cancel, leaving just 2,375kcal.
2. 1 Liter (L) of water at room temperature (20˚C) is put in a fridge with a temperature of 7˚C. How much heat, in kcal, must be removed from the water for it to reach 7˚C?
Step 1- write what you know.
m=1L (or 1kg), ∆T=13˚C, c=1kcal/kg˚C, H=?
The kg's and ˚C's cancel, leaving 13kcal.
As this graph shows, when water reaches 0˚C, it stays there until it is completely melted. The same thing happens when liquid water vaporizes at 100 ˚C.
When a substance goes through phase change the heat energy does into separating molecules not raising the molecular kinetic energy.
This heat associated with phase change is called latent heat.
The amount of heat needed to change 1kg of a substance from solid to liquid at the same temperature (going from melting to melted) is the latent heat of fusion.
The heat required to go from solid to liquid at the melting point can be found by multiplying mass x latent heat of fusion (Lf)
The amount of heat needed to change 1kg of a substance from a liquid to a gas at the same temperature (going from evaporating to evaporated) is the latent heat of vaporization.
The heat required to go from liquid to gas at the boiling point can be found by multiplying mass x latent heat of vaporization (Lv)
Lf = 80kcal/kg = 3.35x105J/kg
Lv = 540kcal/kg = 2.26x106J/kg
This means that it takes 80x more energy to melt 1kg of ice at 0˚C (latent heat) than to raise the temperature of 1kg of water by 1˚C (specific heat). also, it takes 540x more energy to vaporize water at 100˚C (latent) than to raise the temperature by 1˚C (specific).
This all means that it takes 7x more energy to change 1kg of 100˚C water to steam than to change 1kg of ice at 0˚C to water.
1. How much heat does it take to change 0.17kg ice at 0˚C to water at 15˚C?
Step 1- find the latent heat of fusion needed to melt the ice.
cancel the kg's to get 13.6kcal.
Step 2- find the heat needed to raise the temperature of the water to 15˚C.
after canceling the kg's and the ˚C's you are left with 2.55kcal
Step 3- add the two answers together to get the total heat that it took.
13.6kcal + 2.55kcal = 16.15kcal
Here's something to think about:
When you are sweaty you stand in front of a fan to cool off. The air provided by the fan is actually the same temperature as the rest of the air in the room, but the motion promotes evaporation by carrying away molecules. The evaporation of the sweat has a cooling effect on the skin because energy was lost.