# Heat Transfer

1. Melt some ice in water, and measure how much the water’s temperature changes to determine the heat energy released by melting ice (its latent heat of fusion).
2. (optional) Determine the specific heat of an unknown metal, and use that to identify it.

## Resources

• Calorimeter
• Thermometer
• Ice
• Scale
• Stopwatch (www.onlinestopwatch.net)
• Steam Generator
• Small cylinder of an unknown metal

## Background

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For many substances, the relationship between the amount of heat energy $Q$ added or lost per unit mass is proportional to the temperature change. This is written as $$Q=mc\Delta T$$ where $m$ is the mass and $c$ the specific heat for the substance. Some common specific heats are:

• Water: $c$ = 4.184 joules/gram/°C
• Aluminum: $c$ = 0.897 joules/gram/°C
• Copper: $c$ = 0.385 joules/gram/°C
• Iron: $c$ = 0.412 joules/gram/°C

The SI unit of heat is the same as energy – the Joule. Heat can also be measured in calories. One calorie is equal to 4.184 Joules. When a substance changes phase (from liquid to gas, solid to liquid, etc.) some heat energy must be lost or gained in order to effect the change. For example, when ice melts some heat energy must be used to break the intermolecular bonds that give ice its rigid structure. This energy does not go into raising the temperature of the ice. The heat required to melt a mass $m$ of ice is $$Q_{fusion}=mL$$ where the latent heat of fusion of ice is $L$ = 333.55 joules/gram.

In this experiment, you will use a calorimeter, which is essentially just an insulated container. An ideal calorimeter allows no heat to escape, so that any exchanges in thermal energy occur only between the contents of the container. In practice, some heat exchange between the external environment occurs over time, but this can often be ignored if the experiment is arranged carefully.

An ice cube placed in lukewarm water will slowly melt as heat from the water transfers to it. The calorimeter has an inner metal cup for keeping fluids from coming into contact with the insulation, and heat exchange between this cup and the contents does need to be accounted for. As the ice melts, it will convert into an equal mass of water that will continue to be warmed by the surrounding water until the temperatures equalize, and this additional mass of water must also be accounted for.

The total heat change for melting an ice cube will be $$Q_{H_2O}+Q_{cup}+Q_{ice}+ Q_{fusion} = 0$$ $$Q=mc\Delta T$$ $$Q_{fusion}=mL$$