
An unknown metal of mass \[192g\] heated to a temperature of ${100^ \circ }C$ was immersed into a brass calorimeter of mass $128g$ containing $240g$ of water a temperature of ${8.4^ \circ }C$. Calculate the specific heat of the unknown metal if water temperature stabilises at ${21.5^ \circ }C$. (specific heat of brass is $394\,JK{g^{ - 1}}{K^{ - 1}}$)
A. $1232\,JK{g^{ - 1}}K - 1$
B. $458\,JK{g^{ - 1}}{K^{ - 1}}$
C. $654\,JK{g^{ - 1}}{K^{ - 1}}$
D. $916\,JK{g^{ - 1}}{K^{ - 1}}$
Answer
161.1k+ views
Hint: Here in this question, we have to find the specific heat for which we must know that temperature affects the specific heat. According to the thermal property, heat lost by one substance is equal to the heat gain by other substances. This concept shall be used to answer this question.
Formula used:
The formula of heat is,
$Q=mc \bigtriangleup T$
Here, $m$ is the mass, $c$ is the specific of heat of a substance and $\bigtriangleup T$ is the change in temperature.
Complete step by step solution:
As before starting the solution we just write the given data from the question:
Mass of Unknown Metal $m_1$= \[192g\] heated at the temperature of $T_1 ={100^ \circ }C$
Specific heat of unknown metal, $c_1=?$
That is immersed into a brass calorimeter whose mass is $m_2 = 128g$.
Specific heat of brass is $c_2= 394\,JK{g^{ - 1}}{K^{ - 1}}$.
Temperature of water and calorimeter is $T_2={8.4^ \circ }C$
To find the specific heat when the temperature stabilises at ${21.5^ \circ }C$ i.e, final temperature $T_3={21.5^ \circ }C$
Now, by using above all data let’s start the solution of the question,
As we know that, by using thermal property
Heat lost = Heat gained
$m_1 c_1 (T_1-T_3) = m_2 c_2 (T_3-T_2) + m_3 c_3 (T_3-T_2) \\ $
$\Rightarrow 192 \times c_1 \times (100 - 21.5) = 128 \times 394 \times (21.5 - 8.4) + 240 \times 4200 \times (21.5 - 8.4) \\ $
$\Rightarrow c_1= \dfrac{660659.2+13204800}{15072} \\ $
$\Rightarrow c_1 =\dfrac{13865459.2}{15072} \\ $
$\therefore c_1= 916\,JK{g^{ - 1}}{K^{ - 1}}$
Hence, the specific heat of the unknown metal if water temperature stabilises at ${21.5^ \circ }C$ is $916\,JK{g^{ - 1}}{K^{ - 1}}$.
Therefore, the correct answer is D.
Note: The thermal properties of matter is one by which matter’s heat conductivity can be specified or in simple words we can say that thermal properties are those who decide the nature of matter where the heat is present. As for which it is very important to study in depth.
Formula used:
The formula of heat is,
$Q=mc \bigtriangleup T$
Here, $m$ is the mass, $c$ is the specific of heat of a substance and $\bigtriangleup T$ is the change in temperature.
Complete step by step solution:
As before starting the solution we just write the given data from the question:
Mass of Unknown Metal $m_1$= \[192g\] heated at the temperature of $T_1 ={100^ \circ }C$
Specific heat of unknown metal, $c_1=?$
That is immersed into a brass calorimeter whose mass is $m_2 = 128g$.
Specific heat of brass is $c_2= 394\,JK{g^{ - 1}}{K^{ - 1}}$.
Temperature of water and calorimeter is $T_2={8.4^ \circ }C$
To find the specific heat when the temperature stabilises at ${21.5^ \circ }C$ i.e, final temperature $T_3={21.5^ \circ }C$
Now, by using above all data let’s start the solution of the question,
As we know that, by using thermal property
Heat lost = Heat gained
$m_1 c_1 (T_1-T_3) = m_2 c_2 (T_3-T_2) + m_3 c_3 (T_3-T_2) \\ $
$\Rightarrow 192 \times c_1 \times (100 - 21.5) = 128 \times 394 \times (21.5 - 8.4) + 240 \times 4200 \times (21.5 - 8.4) \\ $
$\Rightarrow c_1= \dfrac{660659.2+13204800}{15072} \\ $
$\Rightarrow c_1 =\dfrac{13865459.2}{15072} \\ $
$\therefore c_1= 916\,JK{g^{ - 1}}{K^{ - 1}}$
Hence, the specific heat of the unknown metal if water temperature stabilises at ${21.5^ \circ }C$ is $916\,JK{g^{ - 1}}{K^{ - 1}}$.
Therefore, the correct answer is D.
Note: The thermal properties of matter is one by which matter’s heat conductivity can be specified or in simple words we can say that thermal properties are those who decide the nature of matter where the heat is present. As for which it is very important to study in depth.
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