13. 2kg of substance A at 30°C with specific heat capacity 800Jkg °C is mixed with 4kg of substance B at 60°C with specific heat capacity 500Jkg °C. Find the maximum temperature the mixture will reach if there is no heat exchange with the surroundings.

To find the maximum temperature the mixture will reach, we need to apply the principle of conservation of energy. Since there is no heat exchange with the surroundings, the heat lost by the hotter substance will be equal to the heat gained by the cooler substance.

Let the maximum temperature the mixture will reach be \( T_f \) (in °C).

### Step-by-step solution:

1. **For substance A**:
   - Mass, \( m_A = 2 \) kg
   - Initial temperature, \( T_A = 30 \)°C
   - Specific heat capacity, \( c_A = 800 \) J/kg°C

2. **For substance B**:
   - Mass, \( m_B = 4 \) kg
   - Initial temperature, \( T_B = 60 \)°C
   - Specific heat capacity, \( c_B = 500 \) J/kg°C

3. **Heat lost by substance B (the hotter substance)**:
   - The heat lost is given by: 
     \[
     Q_B = m_B c_B (T_B - T_f)
     \]

4. **Heat gained by substance A (the cooler substance)**:
   - The heat gained is given by: 
     \[
     Q_A = m_A c_A (T_f - T_A)
     \]

5. **Equating the heat lost by B to the heat gained by A** (since no heat is lost to the surroundings):
   \[
   m_B c_B (T_B - T_f) = m_A c_A (T_f - T_A)
   \]

Substitute the given values:
\[
4 \times 500 \times (60 - T_f) = 2 \times 800 \times (T_f - 30)
\]

This simplifies to:
\[
2000 \times (60 - T_f) = 1600 \times (T_f - 30)
\]

Expand both sides:
\[
120000 - 2000T_f = 1600T_f - 48000
\]

Combine like terms:
\[
120000 + 48000 = 1600T_f + 2000T_f
\]
\[
168000 = 3600T_f
\]

Solve for \( T_f \):
\[
T_f = \frac{168000}{3600} = 46.67°C
\]

### Final Answer:
The maximum temperature the mixture will reach is approximately **46.67°C**.

---

Do you have any questions or would you like further details?

Here are some related questions to deepen your understanding:

1. How does the specific heat capacity affect the final temperature in a mixture?
2. What would happen if there was heat exchange with the surroundings?
3. How would the solution change if the masses of the substances were different?
4. What is the physical meaning of the final temperature in terms of energy conservation?
5. How would the mixture temperature change if substance A were initially at a higher temperature than substance B?

**Tip**: Always ensure the units of specific heat capacity are consistent when solving thermodynamics problems.

2kg of substance A at 30°C with specific heat capacity 800 J/kg°C is mixed with 4kg of substance B at 60°C with specific heat capacity 500 J/kg°C. Find the maximum temperature the mixture will reach if there is no heat exchange with the surroundings.

This detailed solution explains how to find the maximum temperature of a mixture involving 2kg of substance A at 30°C and 4kg of substance B at 60°C using specific heat capacities. Learn key thermodynamic concepts like conservation of energy and heat transfer. Suitable for students in Grades 9-12.

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