The reading would remain at 0°C until the ice has melted.  Once the ice is gone, then the thermometer would climb.

Once at 100°C it would stay there while the water boiled away.

• a to b is where the material is a solid.
• b to c is where the solid melts
• c to d is where the liquid is heated
• d to e is where boiling occurs
• e to f shows the resulting gas being "superheated" 

  ---


The mysterious energy that is added without changing the state of the material is called latent heat.  This energy goes into overcoming molecular forces in the material.

How much energy?  It depends on the substance, of course!

To melt a substance, you must add the "heat of fusion" and to boil it, you must add the "heat of vaporization".  This is typically given as "L".  For water, L_fusion=3.34E5J/Kg and L_vap=22.6E5J/Kg

To find this amount of energy, use this equation: Q=mL

AN OLD PROBLEM MIGHT LOOK LIKE THIS:

How much heat does it take to get 2Kg of water from 90°C to 110°C?

Your answer would have been Q=mcDT=(2Kg)(4190J/Kg°C)(110-90°C)=167600J

However, now we see that we are wrong, because the temps given overlap the phase change for water at 100°C.  We must heat the water to 100°C, then boil it, and continue to heat it to 110°C.

Let's break the problem into steps...

• heat water from 90°C to boiling point (Q=mcDT, remember c_water=4190J/Kg°C) =83800J
• boil water (Q=mL) =4520000J
• heat steam to 110°C (Q=mcDT, remember c_steam=2010J/Kg°C)=40200J
• add the 3 answers=4644000J