Name Experiment 25 Report Sheet Calorimetry Desk No. A. Specific Heat of a Metal Unknown No. Trial 1 Trial 2 1. Mass of metal (g) 15.002 22.798 2. Temperature of metal (boiling water) (°C) 100 100 3. Mass of calorimeter (g) 27.932 28.285 4. Mass of calorimeter + water (g) 47.832 48.285 5. Mass of water (g) 20 20 6. Temperature of water in calorimeter (°C) 23.50 24.50 7. Maximum temperature of metal and water from-grapl? (°C) 30.52 34.51 8. Instructor’s approval of graph Calculations for Specific Heat of a Metal 1. Temperature change of water, ΔT (°C) 2. Heat gained by water (J) 3. Temperature change of metal, ΔT (°C) 4. Specific heat of metal, equation 25.5 (J/g°C) 5. Average specific heat of metal (J/g°C) *Show calculations for Trial 1 using the correct number of significant figures. Data Analysis
The Correct Answer and Explanation is:
A. Specific Heat of a Metal
| Measurement | Trial 1 | Trial 2 |
|---|---|---|
| 1. Mass of metal (g) | 15.002 | 22.798 |
| 2. Temperature of metal (°C) | 100.00 | 100.00 |
| 3. Mass of calorimeter (g) | 27.932 | 28.285 |
| 4. Mass of calorimeter + water (g) | 47.832 | 48.285 |
| 5. Mass of water (g) | 20.000 | 20.000 |
| 6. Temperature of water (°C) | 23.50 | 24.50 |
| 7. Max temperature (from graph) (°C) | 30.52 | 34.51 |
Calculations – Trial 1
1. Temperature change of water (ΔT_water):
= Final temp – Initial temp
= 30.52°C – 23.50°C = 7.02°C
2. Heat gained by water (q_water):
= mass × specific heat × ΔT
= 20.000 g × 4.184 J/g°C × 7.02°C
= 587.4 J
3. Temperature change of metal (ΔT_metal):
= 100.00°C – 30.52°C = 69.48°C
4. Specific heat of metal (c_metal):
From heat lost by metal = heat gained by water
qmetal=m⋅c⋅ΔT⇒c=qwaterm⋅ΔTq_{metal} = m \cdot c \cdot \Delta T \Rightarrow c = \frac{q_{water}}{m \cdot \Delta T}qmetal=m⋅c⋅ΔT⇒c=m⋅ΔTqwater c=587.4 J15.002 g×69.48 °C=587.41042.6≈0.5635 J/g°Cc = \frac{587.4 \, \text{J}}{15.002 \, \text{g} \times 69.48 \, \text{°C}} = \frac{587.4}{1042.6} \approx \mathbf{0.5635 \, \text{J/g°C}}c=15.002g×69.48°C587.4J=1042.6587.4≈0.5635J/g°C
5. Average Specific Heat:
Trial 1: 0.5635 J/g°C
Trial 2:
ΔT_water = 34.51 – 24.50 = 10.01°C
q_water = 20 × 4.184 × 10.01 = 837.8 J
ΔT_metal = 100 – 34.51 = 65.49°C
c = 837.8 / (22.798 × 65.49) ≈ 837.8 / 1493.3 ≈ 0.5610 J/g°C
Average Specific Heat = (0.5635 + 0.5610) / 2 = 0.5623 J/g°C
Data Analysis
This experiment aims to determine the specific heat capacity of a metal by using the principle of calorimetry. In both trials, a heated metal sample was placed into a calorimeter containing a known mass of water. By measuring the temperature change in water, we inferred the amount of heat the metal lost, assuming no heat was lost to the surroundings.
In Trial 1, the water temperature increased by 7.02°C, absorbing 587.4 joules of heat. The metal cooled from 100.00°C to 30.52°C, a change of 69.48°C. Using the equation for heat transfer, we calculated the specific heat of the metal as 0.5635 J/g°C. In Trial 2, the result was 0.5610 J/g°C. The average of these values yields a specific heat of 0.5623 J/g°C, which is consistent and indicates experimental reliability.
This value is reasonably close to the specific heat of common metals like iron (0.450 J/g°C) or stainless steel (around 0.50 J/g°C), but the slightly higher value may suggest a metal alloy or experimental error.
Potential sources of error include heat loss to the environment, inaccurate readings of temperature, or delays in transferring the metal to the calorimeter. Using an insulating lid and minimizing time between metal transfer and measurement can improve accuracy. Also, accurate plotting and interpolation of the temperature-time graph are essential for determining the maximum temperature.
This experiment demonstrates energy conservation and helps identify unknown metals based on their thermal properties. The data supports the effectiveness of calorimetry in measuring specific heat, which is fundamental in thermodynamic analysis of substances
