Three liquids P,Q and R are given 4 kg of P at `60^@C and 1kg of R at 50^@C` when mixed produce a resultant temperature `55^@C` . A mixture of 1kg of P at `60^@C` and 1kg of Q at `50^@C` shows a temperature of `55^@C` . What will be the resulting temperature when 1 kg of Q at `60^@C` is mixed with 1 kg of R at `50^@C`?

Answer

To solve the problem, we need to analyze the heat exchange between the liquids using the principle of calorimetry. Let's break it down step by step.

### Step 1: Understand the given data
We have three liquids: P, Q, and R. The following mixtures are given:
1. **4 kg of P at 60°C** and **1 kg of R at 50°C** results in a temperature of **55°C**.
2. **1 kg of P at 60°C** and **1 kg of Q at 50°C** also results in a temperature of **55°C**.

We need to find the resulting temperature when **1 kg of Q at 60°C** is mixed with **1 kg of R at 50°C**.

### Step 2: Set up the equations for the first mixture
For the first mixture:
- Heat lost by P = Heat gained by R

Using the formula \( Q = mc\Delta T \):
- Heat lost by P:
\[ Q_P = 4 \cdot C_P \cdot (55 - 60) = 4C_P \cdot (-5) = -20C_P \]
- Heat gained by R:
\[ Q_R = 1 \cdot C_R \cdot (55 - 50) = C_R \cdot 5 \]

Setting the heat lost equal to the heat gained:
\[ -20C_P + 5C_R = 0 \implies 5C_R = 20C_P \implies C_R = 4C_P \quad \text{(Equation 1)} \]

### Step 3: Set up the equations for the second mixture
For the second mixture:
- Heat lost by P = Heat gained by Q

Using the same formula:
- Heat lost by P:
\[ Q_P = 1 \cdot C_P \cdot (55 - 60) = C_P \cdot (-5) = -5C_P \]
- Heat gained by Q:
\[ Q_Q = 1 \cdot C_Q \cdot (55 - 50) = C_Q \cdot 5 \]

Setting the heat lost equal to the heat gained:
\[ -5C_P + 5C_Q = 0 \implies 5C_Q = 5C_P \implies C_Q = C_P \quad \text{(Equation 2)} \]

### Step 4: Relate C_P, C_Q, and C_R
From Equation 1, we have \( C_R = 4C_P \) and from Equation 2, \( C_Q = C_P \).

### Step 5: Set up the equations for the third mixture
Now we need to find the temperature when **1 kg of Q at 60°C** is mixed with **1 kg of R at 50°C**.

Using the same heat exchange principle:
- Heat lost by Q:
\[ Q_Q = 1 \cdot C_Q \cdot (T - 60) = C_Q \cdot (T - 60) \]
- Heat gained by R:
\[ Q_R = 1 \cdot C_R \cdot (T - 50) = C_R \cdot (T - 50) \]

Setting the heat lost equal to the heat gained:
\[ C_Q \cdot (T - 60) = C_R \cdot (T - 50) \]

### Step 6: Substitute C_R and C_Q
Substituting \( C_Q = C_P \) and \( C_R = 4C_P \):
\[ C_P \cdot (T - 60) = 4C_P \cdot (T - 50) \]

Dividing through by \( C_P \) (assuming \( C_P \neq 0 \)):
\[ T - 60 = 4(T - 50) \]

### Step 7: Solve for T
Expanding and rearranging:
\[ T - 60 = 4T - 200 \]
\[ -60 + 200 = 4T - T \]
\[ 140 = 3T \implies T = \frac{140}{3} \approx 46.67°C \]

### Final Result
The resulting temperature when **1 kg of Q at 60°C** is mixed with **1 kg of R at 50°C** is approximately **46.67°C**.

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