Determination Of Specific Heat Capacity Of A Given Solid

The determination of the specific heat capacity of a given solid is a fundamental experiment in physics that helps us understand how different materials respond to heat. Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. This property is crucial in various real-life applications. For instance, in cooking, knowing the specific heat capacity of different metals can help in choosing the best material for cookware, ensuring even heating and efficient cooking. Similarly, in engineering, understanding the specific heat capacity of building materials helps in designing structures that can withstand temperature changes, enhancing their durability and energy efficiency. This experiment not only deepens our grasp of thermal properties but also illustrates the practical importance of material science in everyday life.

JEE Main 2025: Physics Formula | Study Materials | High Scoring Topics | Preparation Guide

JEE Main 2025: Syllabus | Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

NEET 2025: Syllabus | High Scoring Topics | PYQs

This Story also Contains

1. Aim
2. Apparatus
3. Procedure
4. Observations
5. Calculations
6. Result:
7. Precautions:
8. Solved Examples Based on Determination of Specific Heat Capacity of A Given Solid
9. Summary

Aim

To determine the specific heat capacity of a given solid by the method of mixtures.

Apparatus

A hypsometer, calorimeter, stirrer, a lid and outer jacket, given solid in powder form or in small pieces, balance, weight box, two half-degree thermometers, cold water, and clamp stand.

Theory

In hypsometer, the solid is heated uniformly above room temperature up to a fixed temperature and then solid is added to cold water in the calorimeter.
Heat lost by solid = Heat gained by the water and calorimeter.

Procedure

1. Put two thermometers A and B in a beaker containing water and note their reading. Take one of them, say A to be standard and find the correction to be applied to the other, say B.
2. Put thermometer B in a copper tube of a hypsometer containing the powder of the given solid. Put sufficient water in the hypsometer and place it on a burner.
3. Weigh the calorimeter with a stirrer and lid over it by the physical balance. Record it.

4. Fill about half of the calorimeter with water at about temperature 5⁰C to 8⁰C below room temperature. Now, weigh it again and record it.
5. Heat the hypsometer for about 10 minutes till the temperature of the solid remains steady.
6. Note the temperature of water in the calorimeter. Now, transfer the solid from the hypsometer to the calorimeter quickly. Stir the contents and record the final temperature of the mixture.
7. Remove the thermometer A from the calorimeter and weigh the calorimeter with its contents and lid.

Observations

1. Reading of thermometer A=TA=…0C
2. Reading of thermometer B=TB=…..0C
3. Correction applied in w.r.t A=(TA−TB)=…..0C
4. Mass of calorimeter and stirrer m=…… g
5. Water equivalent of calorimeter =w=m×0.095=…..g
6. Specific heat of copper calorimeter =0.095cal/g
7. Mass of calorimeter + stirrer + lid =m1=…….g
8. Mass of calorimeter + stirrer + lid + cold water =m2=…...g
9. Steady temperature of hot solid =TS=….∘C
10. Corrected temperature of hot solid T=TS−(TA−TB)=….∘C
11. Temperature of cold water =t=…..∘C
12. Temperature of mixture =θ=…..∘C
13. Mass of calorimeter, stirrer,lid, cold water and solid =m3=……g

Calculations

1. Mass of cold wate r=m2−m1=…..g
2. Mass of hot solid =m3−m2=…..g g
3. Rise of the temperature of cold water and calorimeter =θ−t=…..∘C
4. Fall in temperature of solid =T−θ=….∘C
5. Heat gain by calorimeter, cold water and stirrer =[ω+(m2−m1)(θ−t)]=…….
6. Heat lost by solid =(m3−m2)×C×(T−θ)=…….
7. Here, C is the specific heat of the solid to be calculated.

According to the principle of calorimeter, heat lost = heat gained

(m3−m2)×C×(T−θ)=[ω+(m2−m1)(θ−t)]C=[ω+(m2−m1)(θ−t)][(m3−m2)(T−θ)]=…… callg ∘C

Result:

The specific heat of a given solid by the method of the mixture is …calg−1c−1

Precautions:

1. Sufficient solid power should be taken to cover the tip of the thermometer properly.
2. Sufficient water should be taken in a hypsometer.
3. Solid should be dropped quickly and gently.
4. The calorimeter should be polished from the outside to avoid excessive radiation losses.
5. The temperature of cold water should not be below the dew point.

Solved Examples Based on Determination of Specific Heat Capacity of A Given Solid

Example 1: Which of the following substances A, B or C has the highest specific heat in temperature vs time graph?

1) A

2) B

3) C

4) All have equal specific heat

Solution:

Determination of the specific heat capacity of a given solid

S=(H+Mm)×ΔtΔT wherein s= specific heat H= Heat cap of calorie metre ΔT= Temperature gain by object Δt= Temperature loss by body m= mass of solid M= mass of calorimeter ΔQΔTΔt=msΔTΔt⇒ΔtΔt=(ΔQΔt)1ms∴ΔTΔtα1s

Since the slope of C is lowest, hence its specific heat is highest.

Hence, the answer is the option (3).

Example 2: An experiment is conducted to determine the specific heat capacity (c) of a metal using the method of mixtures. A known mass of the metal at a temperature T¹ = 100°C is placed in a calorimeter containing a known mass of water at a lower temperature T₂ = 20°C. The final equilibrium temperature T_f is observed to be 30°C. Given that the mass of the metal is m₁ = 0.2 kg and the mass of water is m₂ = 0.5 kg, calculate the specific heat capacity of the metal.

1) 150 J/kg°C

2) 2500 J/kg°C

3) 1100 J/kg°C

4) 1500 J/kg°C

Solution:

Given values:

The initial temperature of the metal (T₁) = 100°C
Initial temperature of water (T₂) = 20°C
Final equilibrium temperature (T_f ) = 30°C
Mass of the metal (m₁) = 0.2 kg
Mass of water (m₂) = 0.5 kg

The heat gained by the metal (Q₁) is equal to the heat lost by the water (Q₂):

Q_{1 =} Q₂

The heat gained by the metal can be calculated using the formula:

Q1=m1⋅c⋅ΔT1
The heat lost by the water can be calculated using the formula:

Q2=m2⋅cwater ⋅ΔT2

where cwater is the specific heat capacity of water.
Since Q1=Q2, we have:

m1⋅c⋅ΔT1=m2⋅cwater ⋅ΔT2
Solve for c :

c=m2⋅cwater ⋅ΔT2m1⋅ΔT1

Step 1:

Calculate the temperature differences:
ΔT1=|Tf−T1=30∘C−100∘C|=70∘CΔT2=Tf−T2=30∘C−20∘C=10∘C
Step 2: Calculate the specific heat capacity of water:

cwater ―=4200 J/kg∘C
Step 3: Substitute the values and calculate c:

c=0.5 kg⋅4200 J/kg∘C⋅10∘C0.2 kg⋅(70)∘C
Step 4: Calculate c:

c=1500 J/kg∘C
The specific heat capacity of the metal is calculated to be 1500 J/kg∘C.

Hence, the answer is the option (4).

Summary

To determine the specific heat capacity of a particular solid material you must measure how much heat is needed to increase the temperature of one kilogram of the substance by one degree Celsius. Basically, you need to put it in front of a calorimeter and heat it up after which it is put into some amount of water whose mass is already known, hence the answer. ко If а calorimeter is used during this process, the temperature change will help calculate heat energy whiсh was transferred from water in the solid material under consideration.