Q problems Lecture Notes
Q=gCpT
Q= heat in Joules or calories
g=grams
Cp= Specific Heat (J/gC or cal/gC)
T=change in temperature
Conversions:
1 Cal=1kcal=1000cal
1cal=4.18J
1kJ=1000J
Heat is a form of energy. It is the kinetic energy of molecules. Temperature is the
measurement of that energy.
1 calorie is the amount of heat energy needed to raise the temperature of 1g of water by 1C
specific heat is the amount of energy needed to raise the temperature of 1g of any substance
by 1C
specific heat is unique to a substance and also to the phase of matter of the
substance
For example the specific heat of water is 1 cal/gC or 4.18J/gC. For ice it is
2.06J/gC and for water vapor 2.02J/gC
To raise the temp of 1g of water from 19C to 20C takes 1 calorie of energy
To raise the temp of 1g of water from 20C to 80C takes 60 calories of energy
To raise the temp of 5g of water from 20C to 80C takes 300 calories of energy
Examples of Q problems:
1. A student needs to cool 250g of water from 70C to 5C. How much energy in
Joules is required to do this task? What is that energy in kJ, calories, Calories?
Q=gCpT
Q=(250g)(4.18J/gC)(65C)
Q=67,925J
Q= 67.925kJ
67,925J * (1cal/4.18J)= 16,250cal
16,250cal *(1Cal/1000cal) = 16.25Cal
2. Find the specific heat of iron in Joules per gram degree Celsius if it takes
27,600J of energy to raise 300g of iron at 200C.
Q=gCpT
Cp=Q/gT
Cp= 27,600J/(300g*200C)
Cp=0.46J/gC