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.
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = ![\frac{1}{\sqrt{10} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Csqrt%7B10%7D%20%7D)
and
![\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }](https://tex.z-dn.net/?f=%5Cfrac%7Bv_1%7D%7Bv%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bh_1%7D%7B6.1%7D%20%7D)
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
![e = \sqrt{\frac{h_1}{6.1} }](https://tex.z-dn.net/?f=e%20%3D%20%5Csqrt%7B%5Cfrac%7Bh_1%7D%7B6.1%7D%20%7D)
![e = \sqrt{\frac{h_2}{h_1} }](https://tex.z-dn.net/?f=e%20%3D%20%5Csqrt%7B%5Cfrac%7Bh_2%7D%7Bh_1%7D%20%7D)
So on
![e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }](https://tex.z-dn.net/?f=e%5En%20%3D%20%5Csqrt%7B%5Cfrac%7Bh_1%7D%7B6.1%7D%20%7D%5Ctimes%20%5Csqrt%7B%5Cfrac%7Bh_2%7D%7Bh_1%7D%20%7D%5Ctimes...%20%5Csqrt%7B%5Cfrac%7Bh_n%7D%7Bh_%7Bn-1%7D%20%7D)
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
Option (c) : 20°C
Explanation:
![t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2}](https://tex.z-dn.net/?f=t%28final%29%20%3D%20%20%5Cfrac%7Bw1%20%5Ctimes%20t1%20%2B%20w2%20%5Ctimes%20t2%7D%7Bw1%20%2B%20w2%7D%20)
T(final) = 500* 10 + 100*70/600 = 20°C
Answer:
Yes, the rocks are made of matter
Explanation:
Let's remember the definition of matter.
Matter is all that has mass and occupies a place in space. Therefore, if we measure the mass of each rock we will know its mass, the other fact is that rocks like any particular body are occupying a place in an empty space.
The opposite of this is antimatter and can its extent be given by the quantum mechanics.