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Q: What does Specific Heat Capacity: Electrical vs Mixing Methods for A-Level Physics cover?A: Compare electrical heating and mixing methods for measuring specific heat capacity.TL;DR Keep Practical Notes Consistent Use our O-Level Physics Experiments hub to align this write-up with the rest of your Paper 3 practice set.
Why Specific Heat Capacity Matters Specific heat capacity c c c
Q = m c Δ T Q = mc\Delta T Q = m c Δ T
Where:
Q Q Q = Energy transferred (J)
Part 19 of 21
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= Mass (kg)
c c c = Specific heat capacity (J kg⁻¹ K⁻¹)
Δ T \Delta T Δ T = Temperature change (K)
But measuring c c c
Method 1: Electrical Heating
The Setup Metal block (aluminum/copper) with holes
Immersion heater (12V, 50W typical)
Thermometer or temperature probe
Power supply with ammeter/voltmeter
Stopwatch
Insulation (polystyrene/foam)
Insert heater in one hole
Thermometer in other hole
Add oil drops for thermal contact
Wrap block in insulation
Connect power supply
Experimental Procedure Record initial temperature T 0 T_0 T 0 Switch on heater at known power
Record temperature every 30s for 10 minutes
Continue recording after heater off for 5 minutes
Measure voltage
V V V and current
I I I regularly
Calculating Heat Capacity Energy supplied: Q = V I t Q = VIt Q = V I t
From Q = m c Δ T Q = mc\Delta T Q = m c Δ T
c = V I t m Δ T c = \frac{VIt}{m\Delta T} c = m Δ T V I t
Aluminum block:
m = 1.00 m = 1.00 m = 1.00 kg
Heating:
V = 12.0 V = 12.0 V = 12.0 V,
I = 4.00 I = 4.00 I = 4.00 A,
t = 300 t = 300 t = 300 s
Temperature rise:
Δ T = 16.0 \Delta T = 16.0 Δ T = 16.0 K
c = 12.0 × 4.00 × 300 1.00 × 16.0 = 900 J kg − 1 K − 1 c = \frac{12.0 \times 4.00 \times 300}{1.00 \times 16.0} = 900 \text{ J kg}^{-1}\text{ K}^{-1} c = 1.00 × 16.0 12.0 × 4.00 × 300 = 900 J kg − 1 K − 1
(Actual value for aluminum: 897 J kg⁻¹ K⁻¹)
Advantages of Electrical Method ✓ Precise energy measurement (±1 percent)Continuous temperature monitoring Works for solids and liquids Easy to repeat and average
Common Sources of Error Heat losses to surroundings (biggest issue)
Heater not fully immersed
Poor thermal contact (air gaps)
Power fluctuations
Temperature probe lag
Method 2: Mixing Method
The Classic Approach
Experimental Procedure Measure mass of calorimeter m cal m_\text{cal} m cal Add cold water , measure mass
m cold m_\text{cold} m cold Record initial temperature T cold T_\text{cold} T cold Heat separate water to ~60°C
Quickly add hot water , stir rapidly
Record maximum temperature T final T_\text{final} T final Measure total mass for
m hot m_\text{hot} m hot
The Energy Balance Heat lost by hot water = Heat gained by cold water + calorimeter
m hot c ( T hot − T final ) = m cold c ( T final − T cold ) + C ( T final − T cold ) m_\text{hot}c(T_\text{hot} - T_\text{final}) = m_\text{cold}c(T_\text{final} - T_\text{cold}) + C(T_\text{final} - T_\text{cold}) m hot c ( T hot − T final ) = m cold c ( T final − T cold ) + C ( T final − T cold )
Where C C C
Finding Unknown Heat Capacity For unknown liquid (not water):
Use water to calibrate calorimeter first
Repeat with unknown liquid
Solve for
c unknown c_\text{unknown} c unknown Heat metal in boiling water
Transfer quickly to calorimeter
Measure temperature rise
Advantages of Mixing Method ✓ Simple equipment Quick results Good for comparing liquids Minimal electrical knowledge needed
Major Error Sources Heat lost during transfer (critical!)
Incomplete mixing
Calorimeter heat capacity uncertainty
Evaporation losses
Temperature stratification
Heat Loss Corrections
The Cooling Correction Graph Both methods suffer from heat losses. Here's how to correct:
Plot temperature vs time
Identify heating period (electrical) or mixing point (mixing)
Extend cooling curve back
Find corrected Δ T \Delta T Δ T
Graphical Method for Electrical Heating During heating: Temperature rises (heating > losses)After heating: Temperature falls (cooling only)
Plot T vs t for entire experiment
Extrapolate cooling line back to mid-heating time
Read corrected final temperature
Typically adds 1-2°C to
Δ T \Delta T Δ T
Newton's Law of Cooling Rate of cooling ∝ \propto ∝
d T d t = − k ( T − T room ) \frac{dT}{dt} = -k(T - T_\text{room}) d t d T = − k ( T − T room )
Minimizing Heat Losses
For Electrical Method Better insulation
Optimal heating rate
Too fast: Large temperature gradients
Too slow: More time for losses
Aim for 10-20°C rise in 5 minutes
Stirring (for liquids)
For Mixing Method Pre-warm calorimeter
Minimize transfer time
Optimal temperature difference
Comparing the Methods
Accuracy Comparison Electrical method typically achieves:
±2 to 3 percent for metals
±3 to 5 percent for liquids
Better for high heat capacity materials
Mixing method typically achieves:
±5 to 10 percent for water
±10 to 15 percent for metals
Worse for poor thermal conductors
When to Use Each High accuracy needed
Testing solids
Time available for setup
Power supply accessible
Advanced Techniques
Continuous Flow Calorimetry For very accurate c c c
Flow liquid at constant rate
Heat with known power
Measure inlet/outlet temperatures
Steady state eliminates container effects
c = P m ˙ Δ T c = \frac{P}{\dot{m}\Delta T} c = m ˙ Δ T P
Where m ˙ \dot{m} m ˙
Differential Scanning Calorimetry Compare unknown to reference:
Heat both at same rate
Measure power difference
Extremely accurate (±0.1 percent)
Detects phase transitions
Bomb Calorimetry For combustion reactions:
Burn sample in oxygen
Measure temperature rise
Account for all heat capacities
Used for food calorie content
Data Analysis Excellence
Electrical Method Analysis Plot 1: Temperature vs Time
Plot 2: Energy vs Temperature
Should be straight line
Intercept reveals losses
Mixing Method Analysis Plot: Temperature vs Time (detailed)
Uncertainty Calculations For electrical method:
δ c c = ( δ V V ) 2 + ( δ I I ) 2 + ( δ t t ) 2 + ( δ m m ) 2 + ( δ T Δ T ) 2 \frac{\delta c}{c} = \sqrt{\left(\frac{\delta V}{V}\right)^2 + \left(\frac{\delta I}{I}\right)^2 + \left(\frac{\delta t}{t}\right)^2 + \left(\frac{\delta m}{m}\right)^2 + \left(\frac{\delta T}{\Delta T}\right)^2} c δc = ( V δ V ) 2 + ( I δ I ) 2 + ( t δ t ) 2 + ( m δ m ) 2 + ( Δ T δ T ) 2
Temperature: ±3 percent
Electrical: ±1 percent
Mass: ±0.1 percent
Time: ±0.5 percent
Common Exam Questions
Q1: "Why is experimental value lower than data book?" Heat lost to surroundings
Incomplete insulation
Temperature gradients in sample
Some energy heats container/heater
Q2: "Suggest improvements to experiment"
Q3: "Compare advantages of each method" Energy input precisely known
No transfer losses
Continuous monitoring possible
Works for poor conductors
Laboratory Best Practices
Pre-Experiment Checks ✓ Calibrate thermometers (ice/boiling water)Check power supply stability Dry all equipment (water affects results)Room temperature stable (no aircon cycling)Practice technique (especially mixing)
During Experiment ✓ Record room temperature Note any disturbances Keep heating rate constant Stir at regular intervals Continue past target temperature
Data Recording Template Time (s) | Temp (°C) | V (V) | I (A) | Notes
---------|-----------|-------|-------|-------
0 | 22.3 | 12.1 | 4.05 | Start
30 | 23.8 | 12.0 | 4.05 | -
60 | 25.3 | 12.0 | 4.04 | -
... | ... | ... | ... | ...
300 | 38.3 | - | - | Heater off
330 | 37.9 | - | - | Cooling
Linking to Theory
Thermal Physics Connections
Energy Conservation Both methods demonstrate:
Energy cannot be created/destroyed
All energy transfers accounted for
Losses explain discrepancies
Your Success Strategy ✓ Choose method wisely based on materialMinimize heat losses with good insulationApply cooling corrections graphicallyCalculate uncertainties throughoutCompare with data book valuesExplain discrepancies scientificallyShow all working clearlyLink to thermal physics principles
Master both methods and you'll handle any heat capacity question confidently. You'll understand why your car engine needs coolant, why water moderates climate, and why metals feel colder than wood - all from measuring how much energy it takes to warm things up.
Specific Heat Capacity: Electrical vs Mixing Methods for A-Level Physics
