Answer: 2. Solution A attains a higher temperature.
Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.
In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.
Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.
<em>We have a formula for such condition,</em>
.....................................(1)
where:
= temperature difference
- c= specific heat of the body
<u>Proving mathematically:</u>
<em>According to the given conditions</em>
- we have equal masses of two solutions A & B, i.e.
![m_A=m_B](https://tex.z-dn.net/?f=m_A%3Dm_B)
- equal heat is supplied to both the solutions, i.e.
![Q_A=Q_B](https://tex.z-dn.net/?f=Q_A%3DQ_B)
- specific heat of solution A,
![c_{A}=2.0 J.g^{-1} .\degree C^{-1}](https://tex.z-dn.net/?f=c_%7BA%7D%3D2.0%20J.g%5E%7B-1%7D%20.%5Cdegree%20C%5E%7B-1%7D)
- specific heat of solution B,
![c_{B}=3.8 J.g^{-1} .\degree C^{-1}](https://tex.z-dn.net/?f=c_%7BB%7D%3D3.8%20J.g%5E%7B-1%7D%20.%5Cdegree%20C%5E%7B-1%7D)
&
are the change in temperatures of the respective solutions.
Now, putting the above values
![Q_A=Q_B](https://tex.z-dn.net/?f=Q_A%3DQ_B)
![m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1](https://tex.z-dn.net/?f=m_A.c_A.%20%5CDelta%20T_A%3Dm_B.c_B%20.%20%5CDelta%20T_B%5C%5C%5C%5C2.0%5Ctimes%20%5CDelta%20T_A%3D3.8%20%5Ctimes%20%5CDelta%20T_B%5C%5C%5C%5C%20%5CDelta%20T_A%3D%5Cfrac%7B3.8%7D%7B2.0%7D%5Ctimes%20%5CDelta%20T_B%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5CDelta%20T_%7BA%7D%7D%7B%5CDelta%20T_%7BB%7D%7D%20%3D%20%5Cfrac%7B3.8%7D%7B2.0%7D%3E1)
Which proves that solution A attains a higher temperature than solution B.