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Calorimetry Virtual Experiment Lab

A full-workflow chemistry investigation. Choose an investigation, state your hypothesis, identify the independent and dependent variables, collect replicated trials, fit a regression line, and derive specific heat, the calorimeter constant, or the enthalpy of combustion with quantified uncertainty.

Choose an Investigation

Investigation A. Specific Heat of an Unknown Metal

How does the temperature drop of a heated metal compare to the temperature rise of the water it is dropped into, and what does the ratio tell us about the metal's specific heat?

Independent Variable

Initial sample temperature T_sample (vary from 60 °C to 100 °C)

Dependent Variable

Final equilibrium temperature T_f of the water + sample mixture

Controlled Variables
  • Sample mass (constant)
  • Water mass and initial temperature (constant)
  • Calorimeter constant C_cal (assumed calibrated)
  • Atmospheric pressure (open coffee-cup)
Hypothesis Prompt

Predict how the difference (T_sample − T_f) compares to (T_f − T_water). What single property of the metal does that ratio reveal?

Expected Result

Plotting (T_sample − T_f) on y and (T_f − T_water) on x yields a straight line through the origin. Slope = (m_water·c_water + C_cal) / (m_sample·c_sample), so c_sample = (m_water·c_water + C_cal) / (m_sample·slope). Copper sits near 0.385 J/(g·°C), aluminum near 0.897 J/(g·°C).

Procedure
  1. Heat the same metal sample to 60, 70, 80, 90, and 100 °C in turn
  2. Drop each into a fresh water bath at the same starting temperature
  3. Measure T_f and tabulate (T_sample − T_f) and (T_f − T_water)
  4. Fit a straight line and read the slope
  5. Compute c_sample from the slope and compare with the accepted value

Setup

g
°C
g
°C
J/°C
J/(g·°C)

Each "Record Trial" applies realistic measurement noise of ±0.2 °C on the thermometer and ±0.05 g on the balance, then appends the trial to the data table.

Current Setup

heated to 95°Cdrop inm=50gT_water = 22.0°Cm_water = 200 g · C_cal = 10.0 J/°C

T_sample − T_f (°C) vs T_f − T_water (°C)

-0.050.230.500.781.05-0.100.200.500.801.10T_f − T_water (°C)T_sample − T_f (°C)

Record at least 2 trials (or load sample data) to see the plot.

Regression & Error Analysis

Record at least 2 trials to compute the regression. For a defensible fit you should collect 6 or more trials across the full range of the IV.

Data Table

(0 rows)
#TrialSample m(g)T sample(°C)Water m(g)T water(°C)T final(°C)T_s − T_f(°C)T_f − T_w(°C)c_sample (trial)(J/(g·°C))
0 / 500
0 / 500
0 / 500

Reference Guide

Investigation Workflow

Every credible calorimetry result comes from more than a single measurement. It comes from a hypothesis, clearly identified variables, replicated trials, a fit line, and an honest error analysis.

  1. State a testable hypothesis
  2. Identify IV, DV, and controlled variables
  3. Record at least 6 replicated trials across the range of the IV
  4. Fit a regression line and inspect residuals
  5. Quote a final value with uncertainty and percent error

Energy Balance

The first law of thermodynamics says heat lost by the hot side equals heat gained by the cold side and by the calorimeter itself:

mscs(TsTf)=mwcw(TfTw)+Ccal(TfTw)m_{s} c_{s} (T_{s} - T_{f}) = m_{w} c_{w} (T_{f} - T_{w}) + C_{\mathrm{cal}} (T_{f} - T_{w})

Solving for the equilibrium temperature gives:

Tf=mscsTs+(mwcw+Ccal)Twmscs+mwcw+CcalT_{f} = \frac{m_{s} c_{s} T_{s} + (m_{w} c_{w} + C_{\mathrm{cal}}) T_{w}}{m_{s} c_{s} + m_{w} c_{w} + C_{\mathrm{cal}}}

With c_water = 4.184 J/(g·°C), measuring T_f lets you solve for any single unknown in the balance.

Coffee-Cup vs Bomb Calorimetry

Coffee-cup calorimetry. Open to the atmosphere at constant pressure. Measures q_p, which equals ΔH for the reaction. Used for specific heat, heats of solution, and heats of neutralization.

Bomb calorimetry. Sealed rigid bomb at constant volume. Measures q_v, which equals ΔU. For most combustion reactions ΔH is close to ΔU because the Δn_gas term is small.

q=(Ccal+mwcw)ΔTq = (C_{\mathrm{cal}} + m_{w} c_{w}) \Delta T

The slope of q versus fuel mass times molar mass gives the molar enthalpy.

Error Analysis

Report each derived value with uncertainty and percent error.

%error=XexpXacceptedXaccepted×100%\%\,\mathrm{error} = \frac{|X_{\mathrm{exp}} - X_{\mathrm{accepted}}|}{|X_{\mathrm{accepted}}|} \times 100\%

Slope uncertainty propagates to derived quantities via the relative-error rule δX/X = δslope/slope. Largest contributors are thermometer resolution (±0.1 to 0.2 °C), heat loss to the surroundings during sample transfer, and incomplete combustion in bomb runs.

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