Three liquids A, B and C having same sepcific heats have masses m,2m and 3m. Their temperaures are 0,20 and `3theta` respectively. ltBrgt Q. What is the temperature of mixture, when A and B mixed?

Options

A. `(5)/(2)theta`
B. `(5)/(3)theta`
C. `(7)/(3)theta`
D. `(13)/(5)theta`

Answer

To find the temperature of the mixture when liquids A and B are mixed, we can follow these steps:

### Step 1: Define the known quantities
- Mass of liquid A, \( m_A = m \)
- Mass of liquid B, \( m_B = 2m \)
- Initial temperature of liquid A, \( T_A = 0^\circ C \)
- Initial temperature of liquid B, \( T_B = 20^\circ C \)
- Specific heat of both liquids, \( s \) (same for both)

### Step 2: Write the heat gained and lost equations
When two bodies are mixed, the heat lost by the hotter body (B) will be equal to the heat gained by the colder body (A).

The heat gained by A can be expressed as:
\[ Q_A = m_A \cdot s \cdot (T_f - T_A) = m \cdot s \cdot (T_f - 0) = m \cdot s \cdot T_f \]

The heat lost by B can be expressed as:
\[ Q_B = m_B \cdot s \cdot (T_B - T_f) = 2m \cdot s \cdot (20 - T_f) \]

### Step 3: Set up the equation for heat balance
Since the heat gained by A is equal to the heat lost by B:
\[ Q_A = -Q_B \]
Substituting the expressions from Step 2:
\[ m \cdot s \cdot T_f = 2m \cdot s \cdot (20 - T_f) \]

### Step 4: Simplify the equation
We can cancel \( m \) and \( s \) from both sides (assuming they are non-zero):
\[ T_f = 2 \cdot (20 - T_f) \]

### Step 5: Solve for \( T_f \)
Expanding the equation:
\[ T_f = 40 - 2T_f \]
Bringing all \( T_f \) terms to one side:
\[ T_f + 2T_f = 40 \]
\[ 3T_f = 40 \]
Dividing both sides by 3:
\[ T_f = \frac{40}{3} \approx 13.33^\circ C \]

### Final Answer
The temperature of the mixture when liquids A and B are mixed is approximately \( 13.33^\circ C \).

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