A liquid of density `0.85 g//cm^(3)` flows through a calorimeter at the rate of `8.0 cm^(2)//s`. Heat is added by means of a 250 W electric heating coil and a temperature difference of `15^@C` is established in steady state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will be

Options

A. `0.6kcal//kgK`
B. `0.3kcal//kgK`
C. `0.5kcal//kgK`
D. `0.4 kcal//kgK`

Explanation

To solve the problem, we need to find the specific heat of the liquid using the provided information. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify Given Data**:
- Density of the liquid, \( \rho = 0.85 \, \text{g/cm}^3 \)
- Flow rate, \( A = 8.0 \, \text{cm}^2/s \)
- Power of the heating coil, \( P = 250 \, \text{W} \)
- Temperature difference, \( \Delta T = 15 \, \text{°C} \)

2. **Convert Units**:
- Convert the density from \( \text{g/cm}^3 \) to \( \text{kg/m}^3 \):
\[ \rho = 0.85 \, \text{g/cm}^3 = 850 \, \text{kg/m}^3 \]

3. **Calculate Mass Flow Rate**:
- The mass flow rate \( \dot{m} \) can be calculated using the formula:
\[ \dot{m} = \rho \cdot A \cdot v \]
- Since the flow rate is given in \( \text{cm}^2/s \), we need to find the velocity \( v \). However, since we are given the area and not the velocity, we can assume that the flow rate \( A \) is effectively the mass flow rate per unit area.
- Thus, we can express the mass flow rate as:
\[ \dot{m} = \rho \cdot A = 850 \, \text{kg/m}^3 \cdot 8 \times 10^{-4} \, \text{m}^2/s = 0.68 \, \text{kg/s} \]

4. **Apply the Heat Transfer Equation**:
- The power supplied to the liquid is equal to the heat absorbed by the liquid:
\[ P = \dot{m} \cdot c \cdot \Delta T \]
- Rearranging for specific heat \( c \):
\[ c = \frac{P}{\dot{m} \cdot \Delta T} \]

5. **Substitute Values**:
- Substitute the known values into the equation:
\[ c = \frac{250 \, \text{W}}{0.68 \, \text{kg/s} \cdot 15 \, \text{°C}} \]
- Calculate \( c \):
\[ c = \frac{250}{0.68 \cdot 15} = \frac{250}{10.2} \approx 24.51 \, \text{J/(kg·°C)} \]

6. **Convert to Appropriate Units**:
- To convert \( c \) from \( \text{J/(kg·°C)} \) to \( \text{kcal/(kg·K)} \):
\[ 1 \, \text{kcal} = 4184 \, \text{J} \]
\[ c \approx \frac{24.51}{4184} \approx 0.00586 \, \text{kcal/(kg·K)} \approx 0.586 \, \text{kcal/(kg·K)} \]

### Final Answer:
The specific heat of the liquid is approximately \( 0.586 \, \text{kcal/(kg·K)} \).