Objectives: To learn about heat energy, and how to solve for it.

What do people do with it?:

• It's used to solve for the amount on heat energy in a substance, or to solve for the amount of heat energy that has been taken away or added.

• When taking away or adding heat, the motion of the molecules is changing.  It is possible to solve for the amount of heat energy either being added or subtracted, while going through phase changes, or staying in the same phase.

• Q=mCt
• Q=mHf

Q=mHv
• Q= amount of energy (J)
• m= mass (kg)
• C= specific heat (J/°C)
• t= change in temperature (°C)
• Hf= heat of fusion (335000 (J/°C))
• Hv= heat of vaporization (2260000 (J/°C))

• 1) 5kg of solid ice starting at -20 °C is heated to 110 °C.  Solve for the total heat energy used to reach this temperature. (Specific heat of ice is 2000J/°C, of water is 4190J/°C, and of steam is 2010J/°C)
• 2) The temperature of a 2kg block of ice is heated to 105°C from -5°C.  Solve for the total amount of heat energy used to heat the ice.
• 3) The temperature of a 5k block of ice is heated to 47°C from an original temperature of -2°C.  How much heat energy is used to heat the ice?

• 1) Q=mCt  Q=(5kg)(2000 J/°C)(20 °C)= 200,000J
•     Q=mHf     Q=(5kg)(335,000 J/°C)= 1,650,000J
•     Q=mCt Q=(5kg)(4190 J/°C)(100°C)= 2,095,000J
•     Q=mHv     Q=(5kg)(2,260,000 J/°C)= 11,300,000J
•     Q=mCt Q=(5kg)(2010J/°C)(10°C)= 100,500J
•     Total: 200,000J+1,650,000J+2,095,000J+11,300,000J+100,500J=
• 15,345,500J


2)    Q= (2kg)(2000J/°C)(5°C)= 20,000J
       Q=(2kg)(335,000J/°C)= 670,000J
       Q=(2kg)(4190J/°C)(100J°C)= 838,000J
       Q=(2kg)(2,260,000 J/°C)= 4,520,000J
       Q=(2kg)(2010J/°C)(5°C)=20,100J
       Total: 20,000J+670,000J+838,000J+4,520,000J+20,100J= 6,081,000J

3)    Q=(5kg)(2000J/°C)(2°C)= 20,000J
       Q=(5kg)(335,000/°C)= 1,650,000J
       Q=(5kg)(4190J/°C)(47°C)= 984,650J
       Total: 20,000J+1,650,000J+984,650J= 2,654,650J