Identifying an Unknown Metal by Calorimetry
Andre is given an unknown metal sample and asked to identify it. He heats 100.0 g of the metal in boiling water until it reaches 90.0 °C, then quickly transfers it into a foam cup holding 200.0 g of water at 20.0 °C. He stirs gently and records the highest temperature reached: 27.0 °C. Reference specific heats, in J/(g·°C): aluminum 0.90, iron 0.45, copper 0.39, lead 0.13, water 4.18. The foam cup absorbs negligible heat. Andre's partner points out that several drops of the boiling water clung to the metal as he moved it, so the metal was not dry when it entered the cup.
1. Calculate the specific heat of the metal and state which metal it most likely is. Show your reasoning.
Read the worked answer
Energy gained by the water equals energy lost by the metal (assuming no losses):
$$q_{water} = mc\Delta T = (200.0\ \text{g})(4.18\ \text{J/g·°C})(27.0 - 20.0\ \text{°C}) \approx 5{,}852\ \text{J}$$
Set that equal to the metal's heat loss:
$$5{,}852 = (100.0\ \text{g})(c)(90.0 - 27.0\ \text{°C}) = 6{,}300c$$
$$c \approx 0.93\ \text{J/(g·°C)}$$
The closest reference value is aluminum (0.90). Iron (0.45) would require roughly twice the heat loss, and copper and lead are far off. The calculated value runs slightly above aluminum, which is a hint that the measurement has a small systematic error.
2. Do the clinging water droplets make Andre's calculated value too high, too low, or leave it unchanged? Explain your reasoning.
Read the worked answer
Too high. Andre assumes every joule absorbed by the cup water came from 100.0 g of metal. But the hot droplets carried in extra thermal energy that he credited to the metal, and their mass was never weighed into his calculation.
Effect on each quantity:
- The cup water's temperature rise (7.0 °C) is inflated, so the calculated heat release is inflated.
- The metal's mass is still taken as 100.0 g, and its temperature drop is still taken as 63.0 °C.
- Dividing an inflated heat by the same mass and temperature change gives an inflated specific heat.
So the true specific heat is somewhat below 0.93, which actually moves the answer closer to aluminum's 0.90 - but Andre cannot know that from his own data, only from the reference table.
3. The final temperature (27 °C) is far closer to the water's starting temperature than to the metal's. Explain why, using the properties of the two materials.
Read the worked answer
The two objects exchange equal amounts of energy, but they do not change temperature equally. What governs the size of a temperature change is mass × specific heat (the heat capacity):
- Water: $200.0 \times 4.18 \approx 836\ \text{J/°C}$
- Metal: $100.0 \times 0.90 \approx 90\ \text{J/°C}$
The water's heat capacity is roughly nine times larger. For the same quantity of energy, its temperature shifts about one-ninth as much as the metal's.
Check: the metal cooled 63 °C, the water warmed 7 °C; 63/7 = 9. So the final temperature must sit close to the starting temperature of the material that resists change most. This is why a hot metal part dropped into a sink of water barely warms the water.