Specific Heat of Metal

Purpose: The purpose of this lab was to identify the unknown metal that we were given by conducting an experiment and using the formula Q=mcΔT. The options of metals (specific heats) that we may have recieved were:

• Water: 4.184J/g・c
• Aluminum: 0.897J/g・c
• Brass: 0.385J/g・c
• Copper: 0.385J/g・c
• Lead: 0.129J/g・c
• Stainless Steel: 0.490J/g・c
• Zinc: 0.390J/g・c

Background: We needed to learn how to use the formula Q=mcΔT.

• m= mass of metal/water
• c= specific heat
• ΔT= temperature change
• Q= heat

Since it is very difficult to accurately measure the metal's heat directly, we had to transfer the heated metal from the boiling water, to a styrofoam cup (holds heat well) which contained room temperature water. The water in the styrofoam cup then cooled down the metal whilst embodying the heat and becoming warmer.
Procedure: 

1. Find mass of metals and put on a string.
2. Weigh styrofoam cup.
3. Fill styrofoam cup with 100mL of distilled room teperature water.
4. Measure temperature of the water.
5. Fill beaker with 150mL of distilled room temperature water.
6. Weigh syrofoam cup with water in it.

1. Heat the beaker of water on a hot plate.
2. Tie string of metals to ring stand, and let sit in heated beaker of water.
3. Let the water in beaker come to a deep boil.
4. Measure the temperature of the boiling water.
5. When the boiling is adequate and steam is being let off, cut the string with metals, and place in the styrofoam cup and place the lid on top.
6. Measure the temperature of the water in the styrofoam cup once the metal has cooled.

Data: 

• Mass of metals alone: 26.55g
• Mass of styrofoam cup: 8.002g
• Mass of styrofoam cup with water: 92.76g
• Temperature of water in beaker: 13.4°c
• Temperature of water in styrofoam cup: 12.9°c
• Temperature of heated water: 85°c
• Temperature of water in strofoam cup with metal: 14.8°c/14.9°c

Analysis:
First we find the specific heat of the water:

• Q=mcΔT
• Q=(92.76g-8.002g)(4.184)(14.8°c-13.4°c)
• Q=(84.758g)(4.184)(1.4°c)
• Q=496.478J

Then we find the specific heat of the metal:

• Q=mcΔT
• 496.478J=(26.55g)c(85°c-14.8°c)
• 496.478J=(26.55g)c(70.2°c)
• c=0.266J/g・c

Conclusion: The substances specific heat is c=0.266J/g・c. This means that my mystery metal was likely lead.