A "flow calorimeter" is used to measure the specific heat of a liquid. Heat is added at a known rate to a stream of the liquid as it passes through the calorimeter at known rate, then a measurement of the resulting temperature difference between the in flow and the out flow points of the liquid stream enables us to compute the specific heat of the liquid. A liquid of density 0.85 `g//cm^(3)` flows through a calorimeter at the rate of 8.0 `cm^(3)//s`. heat is added by means of a 250 watt electric heating coil, and a temperature difference of `15^(@)C` is established in steady state conditions between the in flow and out flow points.
Q. Specific heat of the liquid is

Options

A. `0.59cal//g.^(@)C`
B. `4cal//g.^(@)C`
C. `2cal//g.^(@)C`
D. `9cal//g.^(@)C`

Answer

To find the specific heat of the liquid using the flow calorimeter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Density of the liquid, \( \rho = 0.85 \, \text{g/cm}^3 \)
- Volume flow rate, \( V = 8.0 \, \text{cm}^3/\text{s} \)
- Rate of heat added, \( Q = 250 \, \text{W} \) (which is equivalent to \( 250 \, \text{J/s} \))
- Temperature difference, \( \Delta T = 15^\circ C \)

2. **Convert Heat Rate to Calories:**
- Since \( 1 \, \text{W} = \frac{1}{4.18} \, \text{cal/s} \), we convert the heat added:
\[ Q = \frac{250 \, \text{J/s}}{4.18 \, \text{J/cal}} \approx 59.8 \, \text{cal/s} \]

3. **Calculate Mass Flow Rate:**
- The mass flow rate \( \dot{m} \) can be calculated using the density and volume flow rate:
\[ \dot{m} = \rho \times V = 0.85 \, \text{g/cm}^3 \times 8.0 \, \text{cm}^3/\text{s} = 6.8 \, \text{g/s} \]

4. **Use the Heat Transfer Equation:**
- The equation relating heat added, mass flow rate, specific heat, and temperature difference is:
\[ Q = \dot{m} \cdot C \cdot \Delta T \]
Rearranging for specific heat \( C \):
\[ C = \frac{Q}{\dot{m} \cdot \Delta T} \]

5. **Substitute Values:**
- Now substitute the values into the equation:
\[ C = \frac{59.8 \, \text{cal/s}}{6.8 \, \text{g/s} \cdot 15 \, \text{C}} \]

6. **Calculate Specific Heat:**
\[ C = \frac{59.8}{6.8 \times 15} \approx \frac{59.8}{102} \approx 0.586 \, \text{cal/g}^\circ C \]

7. **Final Result:**
- Rounding to two decimal places, the specific heat of the liquid is approximately:
\[ C \approx 0.59 \, \text{cal/g}^\circ C \]

### Conclusion:
The specific heat of the liquid is \( 0.59 \, \text{cal/g}^\circ C \).

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