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This problem involves heat transfer between a metal ball and water. Here's the given information translated and summarized:

- The water has a volume of 200 L (liters) and an initial temperature of 40°C.
- A metal ball, heated to 100°C, is dropped into the water.
- The final temperature of the water after reaching thermal equilibrium is 45°C.
- The specific heat capacity of the metal ball is given as 0.46 kJ/(kg·K).
- No heat is lost to the surroundings.

We are asked to find the mass \( m \) of the metal ball in kilograms.

### Approach:
1. **Heat Gained by Water**: The water's temperature increases from 40°C to 45°C. The heat gained by the water is given by:
   \[
   Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
   \]
   - \( m_{\text{water}} \) = mass of water = \( 200 \, \text{L} = 200 \, \text{kg} \) (since 1 L of water is approximately 1 kg).
   - \( c_{\text{water}} \) = specific heat of water = \( 4.18 \, \text{kJ/kg·K} \).
   - \( \Delta T_{\text{water}} \) = change in water temperature = \( 45 - 40 = 5 \, \text{°C} \).

2. **Heat Lost by Metal Ball**: The metal ball cools down from 100°C to 45°C. The heat lost by the metal ball is:
   \[
   Q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}
   \]
   - \( m_{\text{metal}} \) = mass of the metal ball (what we are solving for).
   - \( c_{\text{metal}} = 0.46 \, \text{kJ/kg·K} \) (given in the problem).
   - \( \Delta T_{\text{metal}} = 100 - 45 = 55 \, \text{°C} \).

3. **Thermal Equilibrium Condition**: The heat gained by the water is equal to the heat lost by the metal ball:
   \[
   Q_{\text{water}} = Q_{\text{metal}}
   \]
   Substituting the values:
   \[
   m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}
   \]
   \[
   200 \cdot 4.18 \cdot 5 = m_{\text{metal}} \cdot 0.46 \cdot 55
   \]

Let me calculate the mass of the metal ball.The mass of the metal ball is approximately **165.22 kg**.

Would you like a further breakdown of the steps or additional explanations? Here are some related questions you might want to explore:

1. How would the final temperature change if the mass of the metal ball were smaller?
2. What if the specific heat of the metal ball was different? How would that affect the outcome?
3. How does the assumption that no heat is lost to the surroundings simplify the calculations?
4. What is the significance of specific heat capacity in heat transfer problems?
5. How can we use a similar method to solve problems with different materials or fluids?

**Tip:** Always check that units are consistent throughout the problem, especially when working with heat transfer calculations!

A 100°C heated metal ball is placed in 200L of water at 40°C, and the water's temperature rises to 45°C. Find the mass of the metal ball, given that its specific heat capacity is 0.46 kJ/kg·K. Assume no heat is lost to the surroundings.

In this thermodynamics problem, find the mass of a metal ball placed in 200L of water, using heat transfer and specific heat formulas. The ball starts at 100°C, and the water temperature rises from 40°C to 45°C. Suitable for high school students studying thermodynamics and heat transfer.

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