A thermally insulated calorimeter contains water at a 20°C. A metal ball is heated to 100°C and it is added to the calorimeter. The final temperature of the system was 40°C. After that, metal ball was removed and temperature of the calorimeter with water was cold to 20°C. Then Same amount of water was added to the calorimeter. Again the metal ball is heated to 100°C and it is added to the calorimeter. calculate the new final temperature of the system

To solve this problem, we'll apply the principle of conservation of energy. The heat lost by the metal ball will be equal to the heat gained by the water.

First, let's define some variables:
- \( m_w \) = mass of the water initially in the calorimeter
- \( m_m \) = mass of the metal ball
- \( c_w \) = specific heat capacity of water
- \( c_m \) = specific heat capacity of the metal ball

We are given:
- Initial temperature of the water (\( T_{w,i} \)) = 20°C
- Initial temperature of the metal ball (\( T_{m,i} \)) = 100°C
- Final temperature of the system after the first mixing (\( T_f \)) = 40°C

### First Step: Finding the Mass Ratio

The heat lost by the metal ball (Q_m) and the heat gained by the water (Q_w) in the first scenario are given by:
\[ Q_m = m_m c_m (T_{m,i} - T_f) \]
\[ Q_w = m_w c_w (T_f - T_{w,i}) \]

Since the calorimeter is thermally insulated, \( Q_m = Q_w \):
\[ m_m c_m (100 - 40) = m_w c_w (40 - 20) \]
\[ m_m c_m \cdot 60 = m_w c_w \cdot 20 \]
\[ \frac{m_m c_m}{m_w c_w} = \frac{20}{60} = \frac{1}{3} \]

### Second Step: Finding the New Final Temperature

After the water in the calorimeter is cooled back to 20°C, the same amount of water (m_w) is added to the calorimeter. Therefore, the new mass of the water is \( 2m_w \).

Now, the metal ball is heated again to 100°C and added to the calorimeter. Let the new final temperature be \( T' \).

Again, using the principle of conservation of energy:
\[ Q_m' = Q_w' \]

Heat lost by the metal ball:
\[ Q_m' = m_m c_m (100 - T') \]

Heat gained by the water:
\[ Q_w' = 2m_w c_w (T' - 20) \]

Equating the heat lost by the metal ball to the heat gained by the water:
\[ m_m c_m (100 - T') = 2m_w c_w (T' - 20) \]

Using the ratio \( \frac{m_m c_m}{m_w c_w} = \frac{1}{3} \):
\[ \frac{1}{3} (100 - T') = 2 (T' - 20) \]
\[ 100 - T' = 

Learn how to solve a thermodynamics problem involving a calorimeter and heat transfer. Calculate the final temperature of a system after adding a heated metal ball to water in a thermally insulated container. Suitable for college-level physics students.

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