Explain the procedure of finding specific heat of solid experimentally

Answers (1)
S safeer

Apparatus required
A hypsometer, calorimeter, stirrer, a lid and outer jacket, given solid in powder form or in small pieces,
balance, weight box, two half degree thermometer, cold water, 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 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 the copper tube of hypsometer containing the powder of given solid. Put sufficient
water in hypsometer and place it on a burner.
3. Weigh the calorimeter with stirrer and lid over it by the physical balance. Record it.

4. Fill about half of calorimeter with water at about temperature 50C to 80C below room temperature. Now,
weigh it again and record it.
5. Heat the hypsometer about 10 minutes till the temperature of solid remains steady.
6. Note the temperature of water in the calorimeter. Now, transfer the solid from 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. {\text { Reading of thermometer } A} & {=T_{A}=\ldots \ldots \text { 'C }}

2. \\ {\text { Reading of thermometer } B} & {=T_{B}=\ldots \ldots^{\circ} \mathrm{C}} 

3. \\ {\text { Correction applied in } B \text { w.r.t. A }} & {\left(T_{A}-T_{B}\right)=\ldots . . . \text { 'C }}

4.  \\ {\text { Mass of calorimeter and stirrer }} & {m=\ldots \ldots \mathrm{g}}

5. \\ {\text { Water equivalent of calorimeter }} & {\omega=m \times 0.095=\ldots \ldots \mathrm{g}}

6. \\ {\text { Specific heat of copper calorimeter }} & {=0.095 \mathrm{cal} /\mathrm{g}}

7. Mass \ of \ calorimeter \ +\ stirrer $+$ lid $\quad=m_{1}=\ldots \ldots \mathbf{g}
8. Mass\ of \ calorimeter $+$ stirrer $+$ lid $+$ cold\ water $=m_{2}=\ldots \ldots \mathbf{g}
9. Steady \ temperature\ of\ hot\ solid $\quad=T_{S}=\ldots . .^oC
10. Corrected\ temperature\ of\ hot\ solid\ $T=T_{S}-\left(T_{A}-T_{B}\right)=\ldots . .^{\circ} \mathrm{C}$
11. Temperature \ of \ cold\ water $\quad=t=\ldots . .^{\circ} \mathrm{C}$
12. Temperature\ of\ mixture $\quad=\theta=\ldots . .^{\circ} \mathrm{C}$

13. Mass \ of \ calorimeter, stirrer, lid, cold\ water and\ solid $=m_{3}=\ldots \ldots \mathbf{g}$

 

Calculations:
1. Mass of cold wate r$=m_{2}-m_{1}=\ldots . .$ g
2. Mass of hot solid $=m_{3}-m_{2}=\ldots \ldots \mathrm{g}$
3. Rise of the temperature of cold water and calorimeter $=\theta-t=\ldots . .^{\circ} \mathrm{C}$
4. Fall in temperature of solid $=T-\theta=\ldots .^{\circ} \mathrm{C}$

5. Heat gain by the calorimeter, cold water and stirrer $=\left[\omega+\left(m_{2}-m_{1}\right)(\theta-t)\right]=\ldots \ldots(a)$
6. Heat lost by solid $=\left(m_{3}-m_{2}\right) \times C \times(T-\theta)=\ldots \ldots(a)$
7. Here, C is the specific heat of solid to be calculated.
According to the principle of the calorimeter, the heat lost = heat gained

\left(m_{3}-m_{2}\right) \times C \times(T-\theta)=\left[\omega+\left(m_{2}-m_{1}\right)(\theta-t)\right]$ $ \\ C=\frac{\left[\omega+\left(m_{2}-m_{1}\right)(\theta-t)\right]}{\left[\left(m_{3}-m_{2}\right)(T-\theta)\right]}=

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions