Physics Calorimetry Lab
Calorimetry Lab: Determining the Unknown Metal
Purpose: To determine the identity of an unknown metal
Hypothesis: The unknown metal is Copper
Materials:
* Safety glasses * Styrofoam cup * Thread or string * Glass rod * Thermometer * 100mL graduated cylinder * Hot plate * Balance * Unknown metal * 300mL beaker
Procedure: i. Mass of metal was recorded ii. Water was heated on a hot plate in beaker iii. Temperature was recorded when water reached a boil iv. Cold water was put into graduated cylinder v. Mass of water was recorded vi. Cold water was poured into Styrofoam cup vii. Temperature of cold water was recorded viii. Metal was removed from hot water and was placed into cold water ix. Temperature change in cold water was recorded
Observations:
| Cold Water | Metal | M (kg) | 0.117 kg | 0.07062 kg | T1 (C) | 98C | 81C | T2 (C) | 20C | 20C | T (C) | 2 C | -61C | C ( J/kgC) | 4.2 x 103 | ------------------------------------------------- |
Analysis:
EH=mcT mwcwTw= -mmcmTm cm = mwcwTw /-mmTm
= (0.117)(4.2x103)(2)/(-0.07062)(-61) cm = 228.14 ~ 2.3x102 (Silver) The unknown metal was silver
Discussion:
1. I found the specific heat capacity of my unknown metal in my textbook on page 260 because no metals in the chart matched the calculation I got. 2. Two sources of error for this experiment would be that we cannot determine the actual temperature of the inside of the metal so we just have to assume it is even through, and that our water temperatures changed after we recorded them and before we took the metal out, changing the temperature of the metal so our calculations were not exact. 3. A) Water has a higher specific heat capacity
B) Rock would heat up faster (given the same amount of energy) because it takes more heat energy to change waters temperature by one degree Celsius
C) Rock would lose heat faster because it gained heat
Purpose: To determine the identity of an unknown metal
Hypothesis: The unknown metal is Copper
Materials:
* Safety glasses * Styrofoam cup * Thread or string * Glass rod * Thermometer * 100mL graduated cylinder * Hot plate * Balance * Unknown metal * 300mL beaker
Procedure: i. Mass of metal was recorded ii. Water was heated on a hot plate in beaker iii. Temperature was recorded when water reached a boil iv. Cold water was put into graduated cylinder v. Mass of water was recorded vi. Cold water was poured into Styrofoam cup vii. Temperature of cold water was recorded viii. Metal was removed from hot water and was placed into cold water ix. Temperature change in cold water was recorded
Observations:
| Cold Water | Metal | M (kg) | 0.117 kg | 0.07062 kg | T1 (C) | 98C | 81C | T2 (C) | 20C | 20C | T (C) | 2 C | -61C | C ( J/kgC) | 4.2 x 103 | ------------------------------------------------- |
Analysis:
EH=mcT mwcwTw= -mmcmTm cm = mwcwTw /-mmTm
= (0.117)(4.2x103)(2)/(-0.07062)(-61) cm = 228.14 ~ 2.3x102 (Silver) The unknown metal was silver
Discussion:
1. I found the specific heat capacity of my unknown metal in my textbook on page 260 because no metals in the chart matched the calculation I got. 2. Two sources of error for this experiment would be that we cannot determine the actual temperature of the inside of the metal so we just have to assume it is even through, and that our water temperatures changed after we recorded them and before we took the metal out, changing the temperature of the metal so our calculations were not exact. 3. A) Water has a higher specific heat capacity
B) Rock would heat up faster (given the same amount of energy) because it takes more heat energy to change waters temperature by one degree Celsius
C) Rock would lose heat faster because it gained heat