A piece of copper of mass 1kg heated upto temperature 80°C and then dropped in a. glass beaker filled with 500cc of water at 20°C. After some time the final temperature of water is recorded to be 35°C. The specific heat capacity of copper is: (take specific heat capacity of water `= 4200 J// kg^(@) C` and neglect the thermal capacity of beaker)

Options

A. `420 J// kg^(@) C`
B. `700 J // kg^(@) C`
C. `1180 J// kg^(@) C`
D. `1260J//kg^(@) C`

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the copper will be equal to the heat gained by the water.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of copper, \( m_c = 1 \, \text{kg} \)
- Initial temperature of copper, \( T_{c_i} = 80^\circ C \)
- Mass of water, \( m_w = 500 \, \text{cc} = 0.5 \, \text{kg} \) (since 1 cc of water has a mass of 1 g)
- Initial temperature of water, \( T_{w_i} = 20^\circ C \)
- Final temperature of both, \( T_f = 35^\circ C \)
- Specific heat capacity of water, \( s_w = 4200 \, \text{J/(kg} \cdot \text{°C)} \)

2. **Calculate the change in temperature for copper:**
\[ \Delta T_c = T_{c_i} - T_f = 80^\circ C - 35^\circ C = 45^\circ C \]

3. **Calculate the change in temperature for water:**
\[ \Delta T_w = T_f - T_{w_i} = 35^\circ C - 20^\circ C = 15^\circ C \]

4. **Set up the heat transfer equations:**
The heat lost by the copper is equal to the heat gained by the water:
\[ m_c \cdot s_c \cdot \Delta T_c = m_w \cdot s_w \cdot \Delta T_w \]
Where \( s_c \) is the specific heat capacity of copper.

5. **Substitute the known values into the equation:**
\[ 1 \cdot s_c \cdot 45 = 0.5 \cdot 4200 \cdot 15 \]

6. **Calculate the right side of the equation:**
\[ 0.5 \cdot 4200 \cdot 15 = 31500 \, \text{J} \]

7. **Now, we have:**
\[ 45 s_c = 31500 \]

8. **Solve for \( s_c \):**
\[ s_c = \frac{31500}{45} = 700 \, \text{J/(kg} \cdot \text{°C)} \]

### Final Answer:
The specific heat capacity of copper is \( 700 \, \text{J/(kg} \cdot \text{°C)} \).

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