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

Options

A. a. `(5)/(2)theta`
B. b. `(5)/(3)theta`
C. c. `(7)/(3)theta`
D. d. `(13)/(5)theta`

Answer

To solve the problem of finding the temperature of the mixture when liquids A and B are mixed, follow these steps:

### Step 1: Identify the given data
- Mass of liquid A, \( m_A = m \)
- Mass of liquid B, \( m_B = 2m \)
- Temperature of liquid A, \( T_A = \theta \)
- Temperature of liquid B, \( T_B = 2\theta \)

### Step 2: Set up the heat transfer equation
When two bodies are mixed, the heat lost by the hotter body (liquid B) is equal to the heat gained by the colder body (liquid A).

The equation can be set up as:
\[ \text{Heat lost by B} = \text{Heat gained by A} \]

### Step 3: Write the expressions for heat transfer
Using the formula for heat transfer, \( Q = ms\Delta T \), where \( m \) is mass, \( s \) is specific heat (which is the same for both liquids), and \( \Delta T \) is the change in temperature:

- Heat lost by B:
\[ Q_B = m_B \cdot s \cdot (T_B - T) = 2m \cdot s \cdot (2\theta - T) \]

- Heat gained by A:
\[ Q_A = m_A \cdot s \cdot (T - T_A) = m \cdot s \cdot (T - \theta) \]

### Step 4: Equate the heat lost and gained
Setting the heat lost by B equal to the heat gained by A:
\[ 2m \cdot s \cdot (2\theta - T) = m \cdot s \cdot (T - \theta) \]

### Step 5: Simplify the equation
Since \( s \) and \( m \) can be canceled out from both sides, we simplify:
\[ 2(2\theta - T) = T - \theta \]

### Step 6: Distribute and rearrange
Expanding the left side:
\[ 4\theta - 2T = T - \theta \]

Now, rearranging gives:
\[ 4\theta + \theta = T + 2T \]
\[ 5\theta = 3T \]

### Step 7: Solve for T
Now, divide both sides by 3:
\[ T = \frac{5\theta}{3} \]

### Final Answer
The temperature of the mixture when liquids A and B are mixed is:
\[ T = \frac{5\theta}{3} \]

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