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  4. Determination Of Specific Heat Of Given Liquid By Method Of Mixture - Practice Questions & MCQ

Determination Of Specific Heat Of Given Liquid By Method Of Mixture - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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QuestionMEDIUM

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.

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A metal block of mass m_{1} = 0.2 kg and specific heat c_{metal}= 500 J/kg.K is placed in a container of water of mass m2 = 1.0 kg and specific heat c_{water} = 4186 J/kg·K. Both the metal block and water come to a common final temperature Tf = 320 K. Calculate the initial temperature T1 of the metal block.

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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 isM=0.5 \mathrm{~kg}. The specific heat of water isc_{w}=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}.

The specific heat cm of the metal is:

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Concepts Covered - 1

Determination of specific heat of given liquid by method of mixture

Aim:
To determine 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 hypsometer, the solid is heated uniformly above room temperature up to a fixed temperature and then
solid is added to the given liquid in calorimeter.
Heat lost by solid = Heat gained by the water and calorimeter.

Procedure:
1. Put two thermometer 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 copper tube of hypsometer containing the powder of given solid. Put sufficient
given liquid 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 the given liquid at about temperature 5 to 8oC 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 the given liquid 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 calorimeter and weigh the calorimeter with its contents and lid.

Observations: 

1. \text { Reading of thermometer A} {=T_{A}=\ldots \ldots _{}^{0}\textrm{C}

2. \text { Reading of thermometer B} {=T_{B}=\ldots \ldots _{}^{0}\textrm{C} 

3. \text{Correction applied in w.r.t A}= {\left(T_{A}-T_{B}\right)=\ldots . . . _{}^{0}\textrm{C}

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

5. \text { Water equivalent of calorimeter }= {w=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 \ given\ liquid \quad=t=\ldots . .^{\circ} \mathrm{C}
12. Temperature\ of\ mixture \quad=\theta=\ldots . .^{\circ} \mathrm{C}

13. Mass \ of \ calorimeter, stirrer, lid, given\ liquid\ and\ solid =m_{3}=\ldots \ldots \mathbf{g}

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

5. Heat gain by calorimeter, given liquid and stirrer =\left[\omega+\left(m_{2}-m_{1}\right)(\theta-t)\right]=\ldots \ldots
6. Heat lost by solid =\left(m_{3}-m_{2}\right) \times C \times(T-\theta)=\ldots \ldots
7. Here, C is the specific heat of solid to be calculated.
According to principle of calorimeter, 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]}=\ldots \ldots \operatorname{callg}^{\circ} \mathrm{C}

Result:
Specific heat of given liquid by method of mixture is \ldots cal \mathrm{g}^{-1} \mathrm{c}^{-1}

 

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