Determination Of Specific Heat Of Given Liquid By Method Of Mixture

The determination of the specific heat of a given liquid by the method of mixtures is a fundamental experiment in thermodynamics. This method involves mixing a known mass of the liquid at a certain temperature with another substance, typically water, at a different temperature. By measuring the final equilibrium temperature of the mixture, the specific heat of the liquid can be calculated. This experiment is essential in understanding heat transfer and energy conservation.

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This Story also Contains

1. Procedure
2. Observations
3. Calculations
4. Result
5. Solved Examples Based on Determination Of Specific Heat Of Given Liquid By Method Of Mixture
6. Summary

In real life, the specific heat of liquids plays a crucial role in various applications. For instance, it helps in designing heating and cooling systems, such as radiators and air conditioners, where knowledge of how different fluids absorb and release heat is vital. Another example is in culinary arts, where chefs use the specific heat of water to determine cooking times and temperatures for different recipes, ensuring consistent and precise results.

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

Apparatus
A hypsometer, calorimeter, stirrer, a lid and outer jacket, given solid of known specific heat capacity in powder form or in small pieces, balance, weight box, two half-degree thermometer, Liquid whose specific heat capacity is to be measured, clamp stand.

Theory
In a hypsometer, the solid is heated uniformly above room temperature up to a fixed temperature and then solid is added to the given liquid 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 the given liquid 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 given liquid in a hypsometer and placed 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 the given liquid at about temperature 5 to 8^oC 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 the given liquid 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 $\mathrm{A}=T_A=\dots \dots {}^0\mathrm{C}$
2. Reading of thermometer $\mathrm{B}=T_B=\dots ..{}^0\mathrm{C}$
3. Correction applied in w.r.t $\mathrm{A}=(T_A-T_B)=\dots \dots {}^0\mathrm{C}$
4. Mass of calorimeter and stirrer $m=\dots \dots$ g
5. Water equivalent of calorimeter $=w=m\times 0.095=\dots \dots$ g
6. Specific heat of copper calorimeter $=0.095\mathrm{cal}/\mathrm{g}$
7. Mass of calorimeter + stirrer + lid $=m_1=\dots \dots .\mathrm{g}$
8. Mass of calorimeter + stirrer + lid + cold water $=m_2=\dots$.. g
9. Steady temperature of hot solid $=T_S=\dots {.}^{\circ }\mathrm{C}$
10. Corrected temperature of hot solid $T=T_S-(T_A-T_B)=\dots ..{}^{\circ }\mathrm{C}$
11. Temperature of given liquid $=t=\dots .{.}^{\circ }\mathrm{C}$
12. Temperature of mixture $=\theta =\dots .{.}^{\circ }\mathrm{C}$
13. Mass of calorimeter, stirrer, lid, given liquid and solid $=m_3=\dots \dots \mathrm{g}$

Calculations

1. Mass of given liquid $=m_2-m_1=\dots \dots g$
2. Mass of hot solid $=m_3-m_2=\dots$.. $g$
3. Rise of the temperature of given liquid and calorimeter $=\theta -t=\dots {.}^{\circ }\mathrm{C}$
4. Fall in temperature of solid $=T-\theta =\dots .{}^{\circ }\mathrm{C}$
5. Heat gain by calorimeter, given liquid and stirrer $=[\omega +(m_2-m_1)(\theta -t)]=\dots \dots$
6. Heat lost by solid $=(m_3-m_2)\times C\times (T-\theta )=\dots \dots$
7. Here, $C$ is the specific heat of the solid to be calculated.

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

$\begin{array}{rl}& (m_3-m_2)\times C\times (T-\theta )=[\omega +(m_2-m_1)(\theta -t)]\\ & C=\frac{[\omega +(m_2-m_1)(\theta -t)]}{\lfloor (m_3-m_2)(T-\theta )]}=\dots {\text{callg}}^{\circ }\mathrm{C}\end{array}$

Result

$\text{Specific heat of given liquid by method of mixture is}.\text{. .}{\text{calg}}^{-1}{\mathrm{c}}^{-1}$

Solved Examples Based on Determination Of Specific Heat Of Given Liquid By Method Of Mixture

Example 1: Two liquids of equal volume are thoroughly mixed. If their specific heats are $c_1$ and $c_2$ temperature ${\mathrm{\Theta }}_1,{\mathrm{\Theta }}_2$ and densities $d_1,d_2$ respectively. What is the final temperature of the mixture?

1) $\frac{d_1{c}_1{\mathrm{\Theta }}_1+d_2{c}_2{\mathrm{\Theta }}_2}{d_1{\mathrm{\Theta }}_1+d_2{\mathrm{\Theta }}_2}$
2) $\frac{c_1{\mathrm{\Theta }}_1+c_2{\mathrm{\Theta }}_2}{d_1{c}_1+d_2{c}_2}$
3) $\frac{d_1{\mathrm{\Theta }}_1+d_2{\mathrm{\Theta }}_2}{c_1{\mathrm{\Theta }}_1+c_2{\mathrm{\Theta }}_2}$
4) $\frac{d_1{c}_1{\mathrm{\Theta }}_1+d_2{c}_2{\mathrm{\Theta }}_2}{d_1{c}_1+d_2{c}_2}$

Solution:

$\begin{array}{rl}& \mathrm{\Theta }=\frac{m_1{c}_1{\mathrm{\Theta }}_1+m_2{c}_2{\mathrm{\Theta }}_2}{m_1{c}_1+m_2{c}_2}\\ & m_1=d_{1}v,m_2=d_{2}v\\ & \Rightarrow \mathrm{\Theta }=\frac{d_1{c}_1{\mathrm{\Theta }}_1+d_2{\mathrm{\Theta }}_2{c}_2}{d_1{c}_1+d_2{c}_2}\end{array}$Hence, the answer is the option 4.

Example 2: A metal block of mass $m_1=0.5\mathrm{kg}$ and initial temperature $T_1=300\mathrm{K}$ is placed in a container of water of mass $m_2=2.0\mathrm{kg}$ and initial temperature. $T_2=280\mathrm{K}$. After some time, both the metal block and water come to a common final temperature $T_f=290\mathrm{K}$. Given that the specific heat of the metal is $c_{\text{metal}}=400\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$ and the specific heat of water is $c_{\text{water}}=4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$ calculate the specific heat capacity of the metal block.

1) $41860\mathrm{kJ}/\mathrm{kg}\cdot \mathrm{K}$.
2) $-41860\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$.
3) $41860\mathrm{kJ}/\mathrm{kg}\cdot \mathrm{K}$.
4) $21860\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$.

Solution:

Given:

• - Mass of metal block $m_1=0.5\mathrm{kg}$
  - The initial temperature of the metal block $T_1=300\mathrm{K}$
  - Mass of water $m_2=2.0\mathrm{kg}$
  - The initial temperature of the water $T_2=280\mathrm{K}$
  - Final temperature $T_f=290\mathrm{K}$
  - Specific heat of metal $c_{\text{metal}}=400\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$
  - Specific heat of water $c_{\text{water}}=4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$

  Step 1: Calculate the heat gained by the metal block $\left(Q_{\text{metal}}\right)$ :

  $\begin{array}{c}{Q}_{\text{metal}}=m_1\cdot c_{\text{metal}}\cdot (T_f-T_1)\\ Q_{\text{metal}}=0.5\mathrm{kg}\cdot 400\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\cdot (290\mathrm{K}-300\mathrm{K})\end{array}$
  Step 2: Calculate the heat lost by the water $\left(Q_{\text{water}}\right)$ :

  $\begin{array}{c}{Q}_{\text{water}}=m_2\cdot c_{\text{water}}\cdot (T_2-T_f)\\ Q_{\text{water}}=2.0\mathrm{kg}\cdot 4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\cdot (280\mathrm{K}-290\mathrm{K})\end{array}$

Step 3: Since the heat gained by the metal is equal to the heat lost by the water at thermal equilibrium:

$Q_{\text{metal}}=Q_{\text{water}}$

$0.5\mathrm{kg}\cdot 400\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\cdot (290\mathrm{K}-300\mathrm{K})=2.0\mathrm{kg}\cdot 4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\cdot (280\mathrm{K}-290\mathrm{K})$
Step 4: Solve for the specific heat capacity of the metal $\left(c_{\text{metal}}\right)$ :

$c_{\text{metal}}=\frac{2.0\mathrm{kg}\cdot 4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\cdot (280\mathrm{K}-290\mathrm{K})}{0.5\mathrm{kg}\cdot (290\mathrm{K}-300\mathrm{K})}$
Step 5: Calculate Cmetal :

$c_{\text{metal}}=41860\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$
The specific heat capacity of the metal block is $41860\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$.

Hence, the answer is the option (1).

Example 3: A metal block of mass $m_1=0.2\mathrm{kg}$ and specific heat $c_{\text{metal}}=500\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$ is placed in a container of water of mass $\mathrm{m}2=1.0\mathrm{kg}$ and specific heat $C_{\text{water}}=4186$ $\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$. Both the metal block and water come to a common final temperature ${\mathrm{T}}_{\mathrm{f}}=320\mathrm{K}$. Calculate the initial temperature $T_1$ of the metal block.

1) 320 K

2) 340 K

3) 390 K

4) 230 K

Solution:

Given:

Mass of metal block $m_1=0.2\mathrm{kg}$
Specific heat of metal $C_{\text{metal}}=500\mathrm{J}/\mathrm{kg}.\mathrm{K}$
Mass of water $m_2=1.0\mathrm{kg}$
Specific heat of water $C_{\text{water}}=4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}$
Final temperature $T_f=320\mathrm{K}$
Step 1: Calculate the heat gained by the metal block ( $Q_{\text{metal}}$ ) to reach the final temperature $T_f:$

$Q_{\text{metal}}=m_1\cdot c_{\text{metal}}\cdot (T_f-T_1)$
Step 2: Calculate the heat gained by the water (Qwater) to reach the final temperature $T_f$ :

$Q_{\text{water}}=m_2\cdot c_{\text{water}}\cdot (T_f-T_1)$
Step 3: Since both the metal block and water come to the same final temperature T_f, the heat gained by the metal is equal to the heat gained by the water:
$\begin{array}{c}{Q}_{\text{metal}}=Q_{\text{water}}\\ m_1\cdot c_{\text{metal}}\cdot (T_f-T_1)=m_2\cdot c_{\text{water}}\cdot (T_f-T_1)\end{array}$
Step 4: Solve for the initial temperature $T_1$ of the metal block:

$\begin{array}{rl}& m_1\cdot c_{\text{metal}}\cdot (T_f-T_1)=m_2\cdot c_{\text{uater}}\cdot (T_f-T_1)\\ & 0.2\mathrm{kg}\ast 500\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\ast (320\mathrm{K}-T_1)=1.0\mathrm{kg}\ast 4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\ast (320\mathrm{K}-T_1)\end{array}$
Step 5: Solve for $T_1$ :

$\begin{array}{rl}& 0.2\mathrm{kg}\ast 500\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\ast (320\mathrm{K}-{\mathrm{T}}_1)=1.0\mathrm{kg}\ast 4186\mathrm{J}/\mathrm{kg}\cdot \mathrm{K}\ast (320\mathrm{K}-{\mathrm{T}}_1)\\ & 100(320-{\mathrm{T}}_1)=4186(320-{\mathrm{T}}_1)\\ & 100\ast 320-100\ast {\mathrm{T}}_1=4186\ast 320-4186\ast {\mathrm{T}}_1\\ & 32000-100\ast {\mathrm{T}}_1=1339520-4186\ast {\mathrm{T}}_1\\ & 4086\ast {\mathrm{T}}_1=1307520\\ & T_1=\frac{1307520}{4086}\end{array}$

Step 6: Calculate T_1:

_{$T_1\approx 320.33\mathrm{K}$
The initial temperature ${\mathrm{T}}_1$ of the metal block is approximately 320.33 K .}

The initial temperature T₁ of the metal block is approximately 320.33 K.

Hence, the answer is the option (1).

Example 4: In an experiment, a piece of metal is heated to a high temperature and then placed in an insulated container with water. The initial temperature of the water is $T_i={25}^{\circ }\mathrm{C}$ and the initial temperature of the metal is $T_m={150}^{\circ }\mathrm{C}$, After thermal equilibrium is reached, the final temperature of the water and metal mixture is $T_f={30}^{\circ }\mathrm{C}$ The mass of the metal is $m=0.2\mathrm{kg}$, and the mass of water is $M=0.5\mathrm{kg}$. The specific heat of water is $c_w=4200\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$. The specific heat cm of the metal is:

1) $100\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$
2) $250.5\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$
3) $437.5\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$
4) $500\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$

Solution:

The heat lost by the hot metal is equal to the heat gained by the cold water. This can be expressed using the formula:
$m\cdot c_m\cdot (T_m-T_f)=M\cdot c_w\cdot (T_f-T_i)$
Given:

$\begin{array}{rl}m& =0.2\mathrm{kg}\\ T_m& ={150}^{\circ }\mathrm{C}\\ M& =0.5\mathrm{kg}\\ T_i& ={25}^{\circ }\mathrm{C}\\ T_f& ={30}^{\circ }\mathrm{C}\\ c_w& =4200\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}\end{array}$
Solve for cm :

$c_m=\frac{M\cdot c_w\cdot (T_f-T_i)}{m\cdot (T_m-T_f)}$
Substitute the given values:

$c_m=\frac{0.5\mathrm{kg}\cdot 4200\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}\cdot ({30}^{\circ }\mathrm{C}-{25}^{\circ }\mathrm{C})}{0.2\mathrm{kg}\cdot ({150}^{\circ }\mathrm{C}-{30}^{\circ }\mathrm{C})}$
Calculate the value of cm :

$c_m=\frac{21000\mathrm{J}/\mathrm{kg}}{2400\mathrm{K}}=437.5\mathrm{J}/{\mathrm{kg}}^{\circ }\mathrm{C}$

So, the specific heat cm of the metal is approximately.

Hence, the answer is the option (3).

Summary

The way of mixtures is frequently used to find out what the specific heat of a liquid is. The sample liquid with a known mass is mixed with a known substance of specific heat at a known temperature. After mixing, the ultimate equilibrium temperature is noted down. Based on the conservation of energy principle, the quantity of heat lost by the hotter substance will always be equal to that gained by the colder liquid.