50 g of copper is heated to increase its temperature by `10^@C`. If the same quantity of heat is given to `10g` of water, the rise in its temperature is (specific heat of copper`=420J//kg^(@)//C`)

Options

A. `5^@C`
B. `6^@C`
C. `7^@C`
D. `8^@C`

Explanation

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

### Step 1: Write the formula for heat transfer
The heat transfer can be expressed as:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( Q \) is the heat transferred,
- \( m \) is the mass,
- \( s \) is the specific heat capacity,
- \( \Delta T \) is the change in temperature.

### Step 2: Set up the equation for heat transfer
Since the heat gained by water is equal to the heat lost by copper, we can write:
\[ m_{\text{water}} \cdot s_{\text{water}} \cdot \Delta T_{\text{water}} = m_{\text{copper}} \cdot s_{\text{copper}} \cdot \Delta T_{\text{copper}} \]

### Step 3: Substitute the known values
Given:
- Mass of copper, \( m_{\text{copper}} = 50 \, \text{g} = 0.050 \, \text{kg} \)
- Specific heat of copper, \( s_{\text{copper}} = 420 \, \text{J/(kg°C)} \)
- Change in temperature of copper, \( \Delta T_{\text{copper}} = 10 \, °C \)
- Mass of water, \( m_{\text{water}} = 10 \, \text{g} = 0.010 \, \text{kg} \)
- Specific heat of water, \( s_{\text{water}} = 4200 \, \text{J/(kg°C)} \)

Substituting these values into the equation gives:
\[ 0.010 \cdot 4200 \cdot \Delta T_{\text{water}} = 0.050 \cdot 420 \cdot 10 \]

### Step 4: Simplify the equation
Calculating the right side:
\[ 0.050 \cdot 420 \cdot 10 = 21 \, \text{J} \]
Now the equation becomes:
\[ 0.010 \cdot 4200 \cdot \Delta T_{\text{water}} = 21 \]

### Step 5: Solve for \( \Delta T_{\text{water}} \)
Now, isolate \( \Delta T_{\text{water}} \):
\[ \Delta T_{\text{water}} = \frac{21}{0.010 \cdot 4200} \]
Calculating the denominator:
\[ 0.010 \cdot 4200 = 42 \]
Thus:
\[ \Delta T_{\text{water}} = \frac{21}{42} = 0.5 \, °C \]

### Step 6: Final answer
The rise in temperature of the water is:
\[ \Delta T_{\text{water}} = 0.5 \, °C \]