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

Explanation

To find the specific heat of the liquid, we will follow these steps:

### Step 1: Write down the given information
- Density of the liquid, \( \rho = 0.85 \, \text{g/cm}^3 = 0.85 \times 10^{-3} \, \text{kg/cm}^3 = 850 \, \text{kg/m}^3 \)
- Volume flow rate, \( \dot{V} = 8.0 \, \text{cm}^3/\text{s} = 8.0 \times 10^{-6} \, \text{m}^3/\text{s} \)
- Power of the heating coil, \( P = 250 \, \text{W} \)
- Temperature difference, \( \Delta T = 15 \, \text{°C} \)

### Step 2: Calculate the mass flow rate
The mass flow rate \( \dot{m} \) can be calculated using the formula:
\[ \dot{m} = \rho \cdot \dot{V} \]
Substituting the values:
\[ \dot{m} = 850 \, \text{kg/m}^3 \cdot 8.0 \times 10^{-6} \, \text{m}^3/\text{s} = 6.8 \times 10^{-3} \, \text{kg/s} \]

### Step 3: Calculate the heat supplied per second
The heat supplied \( Q \) by the heating coil in one second is equal to the power:
\[ Q = P \cdot t = 250 \, \text{W} \cdot 1 \, \text{s} = 250 \, \text{J} \]

### Step 4: Use the formula for heat transfer
The heat transfer can also be expressed as:
\[ Q = \dot{m} \cdot c \cdot \Delta T \]
Where \( c \) is the specific heat capacity. Rearranging this gives:
\[ c = \frac{Q}{\dot{m} \cdot \Delta T} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:
\[ c = \frac{250 \, \text{J}}{(6.8 \times 10^{-3} \, \text{kg/s}) \cdot (15 \, \text{°C})} \]

### Step 6: Calculate the specific heat capacity
Calculating the denominator:
\[ \dot{m} \cdot \Delta T = 6.8 \times 10^{-3} \cdot 15 = 0.102 \, \text{kg°C} \]
Now substituting this back into the equation for \( c \):
\[ c = \frac{250}{0.102} \approx 2450.98 \, \text{J/(kg°C)} \]

### Step 7: Convert to kilocalories
To convert joules to kilocalories, we use the conversion factor \( 1 \, \text{kcal} = 4180 \, \text{J} \):
\[ c \approx \frac{2450.98}{4180} \approx 0.585 \, \text{kcal/(kg·K)} \]
This can be approximated to:
\[ c \approx 0.58 \, \text{kcal/(kg·K)} \]

### Final Answer
The specific heat of the liquid is approximately \( 0.58 \, \text{kcal/(kg·K)} \).

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